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    PROBLEMSQ#5.1: In the network of the figure, the switch Kis closed at t = 0 with the capacitoruncharged. Find values for i, di/dt and d

    2i/dt

    2at t = 0+, for element values as

    follows:

    V 100 V

    R 1000 C 1 F

    + R

    V C

    -

    Switch is closed at t = 0 (reference time)

    We know

    Voltage across capacitor before switching = VC(0-) = 0 V

    According to the statement under Q#5.1.

    VC(0+) = VC(0-) = 0 V

    V 100

    iC(0+) = i(0+) = = = 0.1 Amp.

    R 1000Element and initial condition Equivalent circuit at t = 0+

    C

    Sc

    Switch

    Drop

    Rise i(0+) Short circuit

    Drop

    Applying KVL for t 0Sum of voltage rise = sum of voltage drop

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    1

    V = iR + idtC

    Differentiating with respect to t

    di i

    R + = 0dt C

    di(0+) -i(0+)

    = [eq. 1]

    dt CR

    By putting the values of i(0+), C & R

    di(0+) -(0.1)

    =

    dt (1 F)(1 k)di(0+)

    = -100 Amp/secdt

    Differentiating eq. 1 with respect to t

    d2i(0+) -di(0+) 1

    =

    dt2

    dt CR

    Putting the corresponding values

    d2i(0+)

    = 100, 000 amp/sec2dt

    2

    Q#5.2: In the given network, Kis closed at t = 0 with zero current in the inductor.Find the values of i, di/dt, and d

    2i/dt

    2at t = 0+ if

    R 10 L 1 H

    V 100 V

    K+ R

    V

    - L

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    Key closed at t = 0

    iL(0+) = iL(0-) = i(0+) = 0 Amp

    According to the statement under Q#5.2:

    Drop

    Rise

    Open circuit Drop

    i(0+)

    Element and initial condition Equivalent circuit at t = 0+

    oc

    Applying KVL for t 0Sum of voltage rise = sum of voltage drop

    Ldi

    V = iR +

    dt

    Ldi

    = ViR

    dt

    di V - iR

    = [eq. 1]dt L

    di(0+) Vi(0+)R

    =

    dt L

    Putting corresponding values

    di(0+) V(0)R

    =

    dt L

    di(0+) V

    =

    dt Ldi(0+) 100

    =

    dt 1

    di(0+)

    = 100 Amp/sec

    dt

    Differentiating [eq. 1]

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    d2i d V iR

    = -

    dt2

    dt L L

    d2i -Rdi

    =

    dt2 Ldtd

    2i(0+) -Rdi(0+)

    =

    dt2

    Ldt

    Putting corresponding values

    d2i(0+)

    = -1, 000 Amp/sec2

    dt2

    Q#5.3: In the network of the figure, Kis changed from position a to b at t = 0. Solve

    for i, di/dt, and d2

    i/dt2

    at t = 0+ ifR 1000 L 1 H

    C 0.1 FV 100 V

    a K

    b R

    V C L

    Equivalent circuit at t = 0+

    b

    sc

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    Applying KVL for t 0Sum of voltage rise = sum of voltage drop

    1 Ldi

    Ri + idt + = 0 [eq. 1]C dt

    Equivalent circuit at t = 0-

    a

    i(0+)

    sc

    V 100

    iL(0+) = iL(0-) = i(0+) = = = 0.1 Amp

    R 1000

    Initial condition:

    VC(0-) = VC(0+) = 0

    Also we know for t 0VR+ VL + VC = 0

    iR + VL + VC = 0

    At t = 0+i(0+)R + VL(0+) + VC(0+) = 0

    (0.1)(1000) + VL(0+) + 0 = 0

    VL(0+) = -100 Volts

    And

    di

    VL = L

    dt

    di VL

    =

    dt L

    di(0+) VL(0+)=

    dt L

    Putting corresponding values

    di(0+)

    = -100 Amp/sec

    dt

    Differentiating [eq. 1] with respect to t

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    Rdi i Ld2i

    + + = 0

    dt C dt2

    Rdi(0+) i(0+) Ld2i(0+)

    + + = 0

    dt C dt2Putting corresponding values

    d2i(0+)

    = 9, 00000 Amp/sec2

    dt2

    Q#5.4: For the network and the conditions stated in problem 4-3, determine the

    values of dv1/dt and dv2/dt at t = 0+.

    R

    K

    V1C1 C2 V2

    V1 2 V

    V2 1 VR 1 C1 1 F

    C2 F

    After switching:

    V1 V2

    Applying KCL at node 1:

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    V1V2 dV1

    + C1 = 0

    R dt

    V1(0+)V2(0+) dV1(0+)

    + C1 = 0

    R dtPutting corresponding values

    2 V1 V dV1(0+)

    + (1 F) = 0

    1 dt

    Simplifying

    dV1(0+)

    = -1 Volt/sec

    dt

    At node 2:V2V1 dV2

    + C2 = 0

    R dt

    V2(0+)V1(0+) dV2(0+)

    + C2 = 0

    R dt

    Putting corresponding values

    1 V2 V dV2(0+)

    + (1/2) = 0

    1 dt

    Simplifying

    dV2(0+)

    = 2 Volt/sec

    dt

    According to KCL

    Sum of currents entering into the junction must equal to the sum of

    the currents leaving the junction

    Q#5.5: For the network described in problem 4.7, determine values of d2v2/dt

    2and

    d3v2/dt

    3at t = 0+.

    NODE

    R1 +

    R2 V2

    V1 C

    -

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    R1 10 R2 20 C 1/20 F

    VC(0-) = VC(0+) = 0 V

    Applying KCL at NODE for t 0V2 dV2 V2V1

    + C + = 0 (1)

    R2 dt R1At t = 0+

    V2(0+) dV2(0+) V2(0+)V1(0+)

    + C + = 0

    R2 dt R1

    V1 = e-t

    Volts

    V2 = VC(0+) = 0 Volts

    0 dV2(0+) 0e-0+

    + C + = 0

    R2 dt R1

    Simplifying

    dV2(0+)= 2 Volt/sec

    dt

    Differentiating eq. 1 with respect to t

    1 dV2 d2V2 1 dV2 dV1

    + C + - = 0

    R2 dt dt2

    R1 dt dt

    1 dV2 1 d2V2 1 dV2 d(e

    -t)

    + + - = 0

    20 dt 20 dt2

    10 dt dt

    1 dV2 1 d2V2 1 dV2 1 d(e

    -t)

    + + - = 0

    20 dt 20 dt2

    10 dt 10 dt

    3 dV2 1 d2V2 1 d(e

    -t)

    + - = 0

    20 dt 20 dt2 10 dt

    d(e-t) = -e

    -t

    3 dV2 1 d2V2

    + + 0.1e-t = 0 (2)

    20 dt 20 dt2

    At t = 0+

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    3 dV2(0+) 1 d2V2(0+)

    + + 0.1e-t(0+)

    = 0

    20 dt 20 dt2

    Simplifying

    d2V2(0+)

    = -8 Volt/sec2

    dt2

    Differentiating eq. 2

    3 d2V2 1 d

    3V2

    + - 0.1e-t

    = 0

    20 dt2

    20 dt3

    At t = 0+

    3 d2

    V2(0+) 1 d3

    V2(0+)+ - 0.1e

    -t(0+)= 0

    20 dt2

    20 dt3

    Putting corresponding values and simplifying

    d3V2(0+)

    = 26 Volts/sec3

    dt3

    Q#5.6: The network shown in the accompanying figure is in the steady state with the

    switch k closed. At t = 0, the switch is opened. Determine the voltage across theswitch, VK, and dVK/dt at t = 0+.

    SOLUTION:

    VK

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    R 1 L 1 H

    C F

    V 2 V

    Equivalent network before switching

    i(0+) sc

    V 2

    iL(0+) = iL(0-) = i(0+) = = = 2 Amp

    R 1

    1

    Also VK= VC = idtC

    dVK i

    =

    dt C

    At t = 0+

    dVK i(0+)=

    dt C

    dVK 2

    =

    dt (1/2)

    dVK= 4 Volts/sec

    dt

    Q#5.7: In the given network, the switch K is opened at t = 0. At t = 0+, solve for thevalues of v, dv/dt, and d

    2v/dt

    2if

    I 10 A

    R 1000 C 1 F

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    V

    Switch is opened at t = 0

    Equivalent circuit at t = 0-

    No current flows through R so

    VC(0-) = VC(0+) = i(0-)R = V(0+)

    Here

    i(0-) = 0

    VC(0-) = VC(0+) = (0)R

    VC(0-) = VC(0+) = 0 Volts

    For t 0; according to KCL at VV dV+ C = I (1)

    R dt

    At t = 0+

    V(0+) dV(0+)

    + C = I

    R dt

    Simplifying

    dV(0+)

    = 107

    Volts/secdt

    Differentiating (1) with respect to t

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    1 dV d2V

    + C = 0

    R dt dt2

    At t = 0+

    1 dV(0+) d2V(0+)+ C = 0

    R dt dt2

    Simplifying

    d2V(0+)

    = -1010

    Volts/sec2

    dt2

    Q#5.8: The network shown in the figure has the switch K opened at t = 0. Solve for

    V, dV/dt, and d2V/dt

    2at t = 0+ if

    I 1 A

    R 100 L 1 H

    V

    Equivalent circuit before switching:

    Because

    iL(0-) = 0 A

    iL(0-) = iL(0+) = 0 A

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    Therefore

    After switching (t = 0+)

    V

    So

    V(0+) = (I)(R)V(0+) = (1)(100)

    V(0+) = 100 Volts

    For t 0Applying KCL at node V

    V 1

    + Vdt = I (1) {v(t)=LdI }R L dt

    Differentiating (1) with respect to t

    1 dV V+ = 0 (2)

    R dt L

    At t = 0+

    1 dV(0+) V(0+)

    + = 0

    R dt L

    Simplifying

    dV(0+)

    = -10, 000 Volts/sec

    dt

    Differentiating (2) with respect to t

    1 d2V 1 dV

    + = 0

    R dt2

    L dt

    At t = 0+

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    1 d2V(0+) 1 dV(0+)

    + = 0

    R dt2

    L dt

    Simplifying

    d2V(0+)

    = 1, 000, 000 Volts/sec2

    dt2

    Q#5.9: In the network shown in the figure, a steady state is reached with the switch

    Kopen. At t = 0, the switch is closed. For the element values given, determine thevalue of Va(0-) and Va(0+).

    Circuit diagram:

    Vb

    Va

    Equivalent circuit before switching:

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    sc

    Req = (10 + 20) 10(10 + 20)(10)

    Req =

    (10 + 20) + 10

    300

    Req =

    40

    Req = 7.5 After simplification Req

    i(0-)

    5 V

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    V

    i(0-) =

    Req

    5

    i(0-) =

    7.5i(0-) = 0.667 Amp.

    i(0-) = iL(0-)

    Va(0-)

    20

    Va(0-) = (5)

    (10 + 20)

    Va(0-) = 3.334 Volts

    Also equivalent network at t = 0+

    t = 0+

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    Vb

    Va

    Applying KCL at node VaVa5 Va - Vb Va

    + + = 0

    10 20 10

    After simplification we get

    Vb = 5Va10 (i)

    Applying KCL at node VbVb - Va Vb5 2

    + + = 0

    20 10 3

    9Vb3Va+ 10 = 0 (ii)Substituting value of Vb from (i) into (ii)

    9[5Va10]3Va + 10 = 0

    45Va903Va + 10 = 0

    42Va80 = 0

    Va(0+) = 1.905 Volts

    Q#5.10: In the accompanying figure is shown a network in which a steady state is

    reached with switch K open. At t = 0, the switch is closed. For the element values

    given, determine the values of Va(0-) and Va(0+).

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    CIRCUIT DIAGRAM:Vb

    Va

    K

    Equivalent circuit before switching

    Req

    VC(0-) = Va(0-) = 0 V

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    Equivalent network at t = 0+

    Vb

    Va

    Applying KCL at node VaVa5 Va - Vb Va

    + + = 0

    10 20 10

    After simplification we get

    Vb = 5Va10 (i)

    At Vb = 5 V

    5 = 5Va10

    Va(0+) = 3 Volts

    Q#5.11: In the network of figure P5-9, determine iL(0+) and iL() for the conditionsstated in problem 5-9.

    Equivalent circuit before switching:

    +

    -

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    sc

    Req = (10 + 20) 10(10 + 20)(10)

    Req =

    (10 + 20) + 10

    300

    Req =

    40

    Req = 7.5 After simplification Req

    i(0-)

    5 V

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    V

    i(0-) =

    Req

    5

    i(0-) =

    7.5i(0-) = 0.667 Amp.

    i(0-) = iL(0-)

    iL(0-) = iL(0+) = 0.667 Amp.

    Equivalent network at t =

    Va

    Applying KCL at node Va iL()Va5 Va Va

    + + = 0

    10 20 10

    After simplification we get

    Va() = 2 VoltsV Va

    iL() = +10 20

    After simplification we get

    iL() = 0.6 Amp.Q#5.12: In the network given in figure p5-10, determine Vb(0+) and Vb() for theconditions stated in Prob. 5-10.

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    At t = 0-, equivalent network is

    VC(0-) = VC(0+) = 5 Volts

    Also equivalent network at t = is

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    Vb

    Va

    According to KCL at VaVa5 Va - Vb Va

    + + = 0

    10 20 10

    After simplification we get

    Va = 0.2Vb+ 2 (i)

    According to KCL at VbVb5 VbVa

    + = 0

    10 20

    After simplification we get3VbVa10 = 0

    Putting the value of Va we get

    Vb() = 4.286 VoltsQ#5.13: In the accompanying network, the switch K is closed at t = 0 with zero

    capacitor voltage and zero inductor current. Solve for (a) V1 and V2 at t = 0+, (b) V1

    and V2 at t = , (c) dV1/dt and dV2/dt at t = 0+, (d) d2V2/dt2 at t = 0+.CIRCUIT DIAGRAM

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    R1

    V1

    L

    C

    V2

    R2

    Equivalent network after switching

    According to statement under Q#5.13

    At t = 0-

    VC(0-) 0 V

    VC(0+) 0 V

    iL(0-) 0 A

    iL(0+) 0 A

    Equivalent network at t = 0+

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    V2(0+) = iL(0+)(R2)

    V2(0+) = (0)(R2)

    V2(0+) = 0 V

    V1(0+) + V2(0+) = VC(0+)Putting corresponding values

    V1(0+) + 0 = 0

    V1(0+) = 0

    Equivalent circuit at t = V1() = iR2R2

    V

    iR2 =R1 + R2

    VR2V1() =

    R1 + R2VC() = ViR1R1

    VR1

    VC() = VR1 + R2

    After simplification we get

    VR2

    VC() =R1 + R2

    Since

    VC() = V1() + V2()VR2

    V2() =R1 + R2

    V1() = VC() - V2()VR2 VR2

    V1() = -R1 + R2 R1 + R2

    V1() = 0 Volts Self justified(c)

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    V1

    V2

    According to KCL at nodeV1VV1 dV1 1

    + C + V1V2)dt = 0 (i)R1 dt L

    Differentiating with respect to t

    1 dV dV1 d2V1 1

    - + C + (V1V2) = 0 (iii)

    R1 dt dt dt2

    L

    According to KCL at node V2V2 1

    + V2V1)dt = 0 (ii)R2 L

    Differentiating with respect to t

    1 dV2 1+ (V2V1) = 0 (iV)

    R2 dt L

    As

    V1 = V1 + V2

    By putting the value of V1 in (iii) & (iv)

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    1 dV d(V1 + V2) d2(V1 + V2) 1

    - + C + (V1 + V2V2) = 0

    R1 dt dt dt2

    L

    1 dV dV1 dV2 d2V1 d

    2V2 1

    - - + C + + (V1) = 0 (V)R1 dt dt dt dt

    2dt

    2L

    1 dV2 1+ (V2V1) = 0 (iV)

    R2 dt L

    1 dV2 1+ (V2(V1 + V2)) = 0 (iV)

    R2 dt L

    1 dV2 1+ (V2V1 - V2) = 0

    R2 dt L

    1 dV2 1+ (V1) = 0 (Vi)

    R2 dt L

    From (V) & (Vi) we can find the values of dV1/dt & dV2/dt.

    (d)

    Refer part C.

    Q#5.14: The network of Prob. 5-13 reaches a steady state with the switch K closed.At a new reference time, t = 0, the switch K is opened. Solve for the quantities

    specified in the four parts of Prob. 5-13.

    (a)

    Equivalent circuit before switching

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    At t = 0-

    V

    iL(0-) =

    R1 + R2VC(0-) = ViR1(0-)R1Here

    iR1(0-) = iL(0-)VC(0-) = ViL(0-)R1

    VR1VC(0-) = V -

    R1 + R2

    VR2

    VC(0-) =

    R1 + R2

    V2 = iR2R2V

    iR2(0-) = iR1(0-) =

    R1 + R2

    VR2

    V2 =

    R1 + R2

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    VR2

    V2(0-) =

    R1 + R2

    V1(0-) + V2(0-) = VC(0-)

    V1(0-) = VC(0-) - V2(0-)

    VR2 VR2

    V1(0-) = -

    R1 + R2 R1 + R2

    V1(0-) = 0 Volts

    Equivalent network after switching

    V1(0+)

    VC(0+)

    V2(0+)

    VC(0-) = VC(0+)

    VR2

    VC(0+) =

    R1 + R2

    V

    iL(0-) = iL(0+) =R1 + R2

    V2(0+) = iL(0+)R2

    VR2

    V2(0+) =

    R1 + R2

    +-

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    V1(0+) + V2(0+) = VC(0+)

    V1(0+) = VC(0+) - V2(0+)

    VR2 VR2

    V1(0+) = -

    R1 + R2 R1 + R2

    V1(0+) = 0 Volts

    (b)

    Equivalent network at t = At t = capacitor will be fully discharged and acts as an open circuit.

    Hence

    VC() = 0 ViL() = 0 AV2() = iL()R2V2() = (0)R2V2() = 0 Volt(c)

    For t 0, the equivalent network is

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    V1

    V2

    V2 V1

    Applying KCL at node V11 dV1(V1V2)dt + C = 0 (i)L dt

    d -V2

    C (V1 + V2) = (ii)

    dt R2

    As

    V1 = V1 + V2

    1 d(V1 + V2)(V1 + V2V2)dt + C = 0L dt

    1 d(V1 + V2)V1dt + C = 0L dt

    Differentiating (i) with respect to t

    1 d2V1 d

    2V2

    V1+ C + C = 0 (iii)

    L dt2

    dt2

    Differentiating (ii) with respect to t

    d2

    -1 dV2

    C (V1 + V2) = (iV)

    dt2

    R2 dt

    After simplification we get the values of dV1/dt & dV2/dt.

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    (d)

    Refer part c.

    Q#5.15: The switch K in the network of the figure is closed at t = 0 connecting the

    battery to an unenergized network. (a) Determine i, di/dt, and d2i/dt

    2at t = 0+. (b)

    Determine V1, dV1/dt, and d2V1/dt2 at t = 0+.

    K

    V0 i

    V1

    Equivalent circuit after switching

    Here

    VC(0-) = VC(0+) = 0 Volt

    iL(0-) = iL(0+) = 0 A

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    i(0+) = 0 A

    iL(0+) = i(0+)

    Applying KVL around outside loop

    di

    L + iR2 = V0 (i)

    dtAt t = 0+

    di(0+)

    L + i(0+)R2 = V0 (i)

    dt

    By putting corresponding values we get

    di(0+) V0

    =

    dt L

    Differentiating (i) with respect to td

    2i di

    L + R2 = 0

    dt2

    dt

    At t = 0+

    d2i(0+) di(0+)

    L + R2 = 0

    dt2

    dt

    Simplifying we get

    d2i(0+) -RV0

    =dt

    2L

    2

    Referring to the network at t = 0+

    V1(0+) = VR1(0+) = iR1(0+)(R1)

    V0

    iR1(0+) =

    R1

    V0R1

    V1(0+) =

    R1

    V1(0+) = V0

    (b)

    Also

    V1 = V0 for all t 0

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    dV1(0+) d2V1(0+)

    = = 0

    dt dt2

    Q#5.16: The network of Prob. 5.15 reaches a steady state under the conditions

    specified in that problem. At a new reference time, t = 0, the switch K is opened.

    Solve for the quantities specified in Prob. 5.15 at t = 0+.

    Equivalent circuit before switching

    V0

    R2

    V0iL(0-) =

    R2

    VC(0-) = VR2(0-) = iL(0-)(R2)

    V0

    VC(0-) = VR2(0-) = (R2)

    R2

    VC(0-) = VR2(0-) = V0

    As

    VC(0-) = VC(0+) = V0

    Equivalent network at t =0+

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    +

    -

    i(0+)

    V0

    i(0+) = iL(0-) = iL(0+) =

    R2V1(0+) = V0VR1(0+)

    V1(0+) = V0iL(0+)R1

    V0

    V1(0+) = V0 R1R2

    R2R1

    V1(0+) = V0

    R2

    Equivalent circuit for t 0

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    R1L

    i

    C

    R2

    Applying KVL around the loop

    1 dii(R1 + R2) + idt + L = 0 (i)

    C dt

    Also

    VR1 + VR2 + VL + VC = 0

    i(0+)R1 + i(0+)R2 + VL(0+) + VC(0+) = 0

    VL(0+) = -i(0+)R1 - i(0+)R2 - VC(0+)

    Here

    VC(0+) = -V0

    V0i(0+) =

    R2V0 V0

    VL(0+) = - R1 - R2(-V0)

    R2 R2

    V0VL(0+) = - R1V0 + V0

    R2

    V0VL(0+) = - R1

    R2

    And

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    di

    VL = L

    dt

    di(0+) VL(0+)

    =

    dt LPutting corresponding value

    di(0+) -V0R1

    =

    dt R2L

    Differentiating eq. (i) with respect to t

    1 di

    i(R1 + R2) + idt + L = 0 (i)C dt

    di i d2i

    (R1 + R2) + + L = 0

    dt C dt2

    At t = 0+

    di(0+) i(0+) d2i(0+)

    (R1 + R2) + + L = 0

    dt C dt2

    Here

    di(0+) -V0R1

    =dt R2L

    V0

    i(0+) =

    R2Putting corresponding values and simplifying

    d2i(0+) V0 R1(R1 + R2) 1

    = -

    dt2

    R2 L C

    Also

    di

    V1 = L + iR2

    dt

    Differentiating with respect to t

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    dV1 d2i di

    = L + R2

    dt dt2

    dt

    At t = 0+

    dV1(0+) d2i(0+) di(0+)

    = L + R2dt dt

    2dt

    By putting corresponding values and simplifying

    dV1(0+) V0 R12

    1

    = -

    dt R2 L C

    We know

    -1

    V1 = idt - iR1C

    Differentiating with respect to t

    dV1 -i di

    = - R1

    dt C dt

    Differentiating with respect to t

    d2V1 -di 1 d

    2i

    = - R1dt

    2dt C dt

    2

    At t = 0+

    d2

    V1(0+) -di(0+) 1 d2

    i(0+)= - R1

    dt2

    dt C dt2

    Putting corresponding values and simplifying

    d2V1(0+) V0R1 2 R1(R1 + R2)

    = -

    dt2

    R2L C L

    Q#5.17: In the network shown in the accompanying figure, the switch K is changed

    from a to b at t = 0. Show that at t = 0+,

    Vi1 = i2 = -

    R1 + R2 + R3

    i3 = 0

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    a C3

    bV R2

    + R3

    i1 i2

    - i3

    R1 L1

    C1 C2

    Equivalent circuit before switching

    At t = 0-, capacitor C3 is fully charged to voltage V that is VC3(0-) = V and behaves

    as an open circuit, so current in L1, L2 becomes and other two capacitors also fully

    charged.

    iL1(0-) = iL1(0+) = 0 A

    iL2(0-) = iL2(0+) = 0 A

    VC1(0-) = VC1(0+) = 0 V

    VC2(0-) = VC2(0+) = 0 V

    Equivalent circuit after switching

    L2

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    After simplification we get

    i2

    i1

    Hence

    -V

    i1 = i2 =

    R1 + R2 + R3C1 behaves short circuit being uncharged at t = 0- & L1 behaving open circuit since

    iL(0-) = iL(0+) = 0 A

    and i3 = 0 [L2 behaving open circuit].

    Q#5.18: In the given network, the capacitor C1 is charged to voltage V0 and the

    switch K is closed at t = 0. When

    R1 2 MV0 1000 V

    + -

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    R2 1 MC1 10 FC2 20 F

    solve for d2i2/dt

    2at t = 0+.

    +

    C1 i1

    V0 C2

    - i2

    R2

    R1

    Equivalent circuit before switching

    i2

    VC1(0-) = V0VC2(0-) = 0 VEquivalent circuit after switching

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    K

    +

    V0 i1 i2

    -

    VC1(0+) = V0VC2(0+) = 0 V

    For t 0For loop 1:

    1

    R2(i1i2) + i1dt = 0 (i)C1

    For loop 2:1

    R2(i2i1) + i2dt + R1i2= 0 (ii)C2

    1

    R2(i1i2) = - i1dtC1

    1

    R2(i2i1) = i1dt (iii)C1

    Taking loop around outside

    i2R1 = V0

    V0

    i2= (a)R1

    In loop 1

    According to KVL:R2(i1i2) = V0R2i1R2i2 = V0Putting the value of i2 and simplifying

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    V0(R1 + R2)

    i1 = (b)

    R1R2

    Putting corresponding values we get

    i1(0+) 1.5(10- ) Amp.

    i2(0+) 5(10-

    ) Amp.

    Substituting value of R2(i2i1) in eq. (ii)

    1 1i1dt + i2dt + R1i2 = 0C1 C2Differentiating with respect to t

    i1 di2 i2

    + R1 + = 0 (iv)

    C1 dt C2

    At t = 0+i1(0+) di2(0+) i2(0+)

    + R1 + = 0

    C1 dt C2By putting corresponding values we get

    di2(0+)

    = -8.75(10-5

    ) Amp/sec.

    dt

    Differentiating eq. (iii) with respect to t

    di2 di1 i1

    R2 -R2 =dt dt C1

    At t = 0+

    di2(0+) di1(0+) i1(0+)

    R2 - R2 =dt dt C1

    By putting corresponding values and simplifying

    di1(0+)

    = -2.375(10-4

    ) Amp/sec.

    dt

    Differentiating eq. (iv) with respect to t

    1 di1 d2i2 1 di2

    + R1 + = 0

    C1 dt dt2

    C2 dt

    At t = 0+

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    1 di1(0+) d2i2(0+) 1 di2(0+)

    + R1 + = 0

    C1 dt dt2

    C2 dt

    Putting corresponding values and simplifying

    d2i2(0+)= 1.40625(10

    -5) Amp/sec

    2.

    dt2

    Q#5.19: In the circuit shown in the figure, the switch K is closed at t = 0 connecting

    a voltage, V0sin t, to the parallel RL-RC circuit. Find (a) di1/dt and (b) di2/dt at t =0+.

    i1 i2

    Equivalent circuit after switching

    At t 0Applying KVL around outside loop

    di2

    +

    -

    +

    -

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    Ri2 + L = V0sin t (i)dt

    Applying KVL around inside loop

    1

    Ri1 + i1dt = V0sin t (ii)C

    Equivalent circuit at t =0+

    iL(0+) = iL(0-) = 0 A

    VC(0+) = 0 V

    V0sin ti1 =

    R

    At t = 0+

    V0sin (0+)i1(0+) =

    R

    V0sin 0

    i1(0+) =

    R

    i1(0+) = 0 A

    From (i)

    At t = 0+

    di2(0+)

    Ri2(0+) + L = V0sin (0+) (i)dt

    By putting corresponding values we get

    di2(0+)

    = 0 Amp/sec

    dt

    Differentiating eq. (ii) with respect to t

    +

    -

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    1

    Ri1 + i1dt = V0sin t (ii)C

    di1 i1

    R + = V0cos tdt C

    At t = 0+

    di1(0+) i1(0+)

    R + = V0cos (0+)dt C

    By putting corresponding values & simplifying we get

    di1(0+) V0=

    dt R

    Q#5.20: In the network shown, a steady state is reached with the switch K open with

    V 100 V

    R1 10 R2 20 R3 20 L 1 H

    C 1 F. At time t = 0, the switch is closed.

    (a)Write the integrodifferential equations for the network after the switch isclosed.(b)What is the voltage V0 across C before the switch is closed? What is itspolarity?

    (c)Solve for the initial value of i1 and i2(t = 0+).(d)Solve for the values of di1/dt and di2/dt at t = 0+.(e)What is the value of di1/dt at ?

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    Circuit diagram:i

    i2 i1

    Equivalent circuit before switching:

    R2

    R1

    R3

    Simplifying

    i

    VC(0-)

    VC(0-) = iR2(R2) = VR2

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    Here

    V

    iR2 =

    R1 + R2

    VR2

    VR2 =R1 + R2

    By putting corresponding values we get

    VC(0-) = VR2 =66.667 V

    V

    iL(0-) =

    R1 + R2

    iL(0-) = 3.334 A

    For t 0

    i2 i1

    Applying KVL around outside loop

    di1

    R2i1 + L = V (i)

    dt

    Applying KVL around inside loop

    1

    R3i2 + i2dt = V (ii)C

    Since

    10

    iL(0+) = iL(0-) = iR2(0-) = i1(0-) = i1(0+) = Amp.

    3

    VVC(0+)

    i2(0+) =

    R3

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    Here

    VC(0+) = 6.667 Volts

    Putting corresponding values & simplifying

    i2(0+) = 1.667 Amp.

    From eq. (i)

    At t = 0+di1(0+)

    R2i1(0+) + L = V (i)

    dt

    Putting corresponding values we get

    di1(0+)

    = 33.334 Amp/sec.

    dt

    Differentiating eq. 2:

    di2 i2

    R3 + = 0

    dt C

    At t = 0+

    di2(0+) i2(0+)

    R3 + = 0

    dt C

    Putting corresponding values

    di2(0+)

    = 83.334(104

    ) Amp/sec.dt

    From eq. (i)

    At t = di1()

    R2i1() + L = V (i)dt

    Here

    di1()= 0 Amp/sec

    dt

    i1() = 5 Amp.

    Equivalent circuit after switching:

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    iL(0+)

    VC(0+)

    Q#5.21: The network shown in the figure has two independent node pairs. If the

    switch K is opened at t = 0, find the following quantities at t = 0+:

    (a) V1(b)V2(c)dV1/dt(d)dV2/dt

    Circuit diagram:V1 V2

    L

    i(t) K R1 C

    R2

    Initial conditions:iL(0-) = iL(0+) = 0 A

    VC(0-) = VC(0+) = 0 V

    Applying KCL at node V11 V1(V1V2)dt + = i(t) (i)L R1

    Differentiating with respect to t

    +-

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    (V1V2) 1 dV1 di(t)

    + =

    L R1 dt dt

    At t = 0+

    (V1(0+)V2(0+)) 1 dV1(0+) di(0+)

    + =L R1 dt dt

    Putting corresponding values we get

    dV1(0+) di(t)(0+) V1(0+) R1

    = -

    dt dt L

    Applying KCL at node V2

    1 V2 dV2(V2V1)dt + + C = 0 (ii)L R2 dtAt t = 0+

    (V2(0+)V1(0+)) 1 dV2(0+) dV2(0+)

    + + C = 0

    L R2 dt dt

    Putting corresponding values we get

    dV2(0+)

    = 0 V/sec.

    dt

    Equivalent circuit at t = 0+ V1(0+) V2(0+)

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    Q#5.22: In the network shown in the figure, the switch K is closed at the instant t =

    0, connecting an unenergized system to a voltage source. Show that if V(0) = V,

    then:

    di1(0+)/dt, di2(0+)/dt =? L1 L2

    R1

    L3

    R3

    i1 i2

    R2

    iL1(0-) = iL1(0+) = 0 A

    iL2(0-) = iL2(0+) = 0 A

    For t 0According to KVL

    Loop 1:

    di1 d(i1i2) d(i1i2) di1 di2

    R1i1 + L1 + L3 + R2(i1i2)+ M13 + M31 + (-M32) = V(t)

    dt dt dt dt dtAfter simplification

    di1 di2

    i(R1 + R2)i2R2 + (L1 + L3 + 2M13) - (L3 + M13 + M23) = V(t) (i)

    dt dt

    i1(0+) = i2(0+) = 0

    At t = 0+

    di1(0+) di2(0+)

    i(0+)(R1 + R2)i2(0+)R2 + (L1 + L3 + 2M13) - (L3 + M13 + M23) = V(t)

    dt dtPutting corresponding values we get

    di1(0+) di2(0+)

    (L1 + L3 + 2M13) - (L3 + M13 + M23) = V (ii)

    dt dt

    According to KVL

    +

    -

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    Loop 2:di2 d(i2i1) d(i2i1) di1 di2

    R3i2 + L2 + L3 + R2(i2i1)+ M23 - M31 + M32 = 0

    dt dt dt dt dt

    After simplifying

    di2 di1

    i2(R3 + R2)i1R2 + (L2 + L3 + 2M23) - (L3 + M13 + M23) = 0 (iii)

    dt dt

    At t = 0+

    di2(0+) di1(0+)

    i2(0+)(R3 + R2)i1(0+)R2 + (L2 + L3 + 2M23) - (L3 + M13 + M23) = 0

    dt dt

    Putting corresponding values we get

    di2(0+) di1(0+)(L2 + L3 + 2M23) - (L3 + M13 + M23) = 0 (iv)

    dt dt

    From (ii) & (iv) we can determine the values of di1(0+)/dt & di2(0+)/dt.

    MASHAALLAH BHAIS REFERENCE:

    In order to indicate the physical relationship of the coils and, therefore, simplify the

    sign convention for the mutual terms, we employ what is commonly called the dot

    convention. Dots are placed beside each coil so that if currents are entering both

    dotted terminals or leaving both dotted terminals, the fluxes produced by these

    currents will add.If one current enters a dot and the other current leaves a dot, the mutual induced

    voltage and self-induced voltage terms will have opposite signs.

    Q#5.23: For the network of the figure, show that if K is closed at t = 0, d2i1(0+)/dt

    2?

    CIRCUIT DIAGRAM:

    L

    R1

    i1 C i2 R2

    V(t)

    +

    -

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    Initial conditions:iL(0-) = iL(0+) = 0 A = i2(0+)

    VC(0-) = VC(0+) = 0 V

    Equivalent circuit after switching

    V(t)

    i1(t) =

    R1

    At t = 0+

    V(0+)

    i1(0+) =

    R1For t 0Loop 1:

    1

    R1i1 + (i1i2)dt = V(t) (i)C

    Differentiating (i) with respect to t

    di1 (i1i2) dV(t)

    R1 + = (ii)

    dt C dt

    At t = 0+

    di1(0+) (i1(0+)i2(0+)) dV(0+)

    R1 + =dt C dt

    Putting corresponding values we get

    +

    -

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    di1(0+) dV(0) V(0) 1

    = -

    dt dt R1C R1

    Differentiating (ii) with respect to t

    d2i1 1 di1 di2 d

    2V(t)

    R1 + - =dt

    2C dt dt dt

    2

    At t = 0+

    d2i1(0+) 1 di1(0+) di2(0+) d

    2V(0+)

    R1 + - =dt

    2C dt dt dt

    2

    From here we can determine the value of d2

    i1(0+)/dt2

    .

    Q#5.24: The given network consists of two coupled coils and a capacitor. At t = 0,

    the switch K is closed connecting a generator of voltage, V(t) = V sin (t/(MC)1/2

    ).

    Show that

    Va(0+) = 0, dVa(0+)/dt = (V/L)(M/C)1/2

    , and d2Va(0+)/dt

    2= 0.

    CIRCUIT DIAGRAM:

    M

    K

    a

    Va

    V(t)

    +

    -

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    Equivalent circuit at t = 0+

    After simplification

    diL(0+)

    Va(0+) = VC(0+) + M

    dt

    We know for t 0, according to KVLdi 1L + idt = V(t) (a)

    dt C

    At t = 0+

    di(0+)

    L + VC(0+) = V(0+)

    dt

    +

    -

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    Here

    VC(0+) = 0 V

    V(t) = V sin (t/(MC)1/2

    )

    V(0+) = V sin ((0+)/(MC)1/2

    )

    V(0+) = V sin (0)

    V(0+) = V (0)V(0+) = 0 V

    Putting corresponding values we get

    di(0+)

    = 0 Amp/sec.

    dt

    iL(0-) = iL(0+) = i(0+) = 0 A

    Now

    di(0+)

    Va(0+) = VC(0+) + M (i)

    dt

    Putting corresponding values we getVa(0+) = 0 Volt

    Now

    Differentiating (i) with respect to t

    dVa dVC d2i

    = + M (b)

    dt dt dt2

    Differentiating (a) with respect to t

    d2i i dV(t)

    L + = (a)

    dt2

    C dt

    HereV

    d(V(t)) = cos (t/(MC)1/2

    )

    (MC)1/2

    At t = 0+

    d2i(0+) i(0+) dV(0+)

    L + = (c)

    dt2

    C dt

    Putting corresponding values we get

    d2i(0+) V

    =

    dt2

    L(MC)1/2

    At node a, apply KCL

    1 dVC(VCV(t)) + C = 0 (c)L dt

    Rearranging

    dVC(0+)

    iL(0+) + C = 0

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    dt

    dVC(0+)

    0 + C = 0

    dt

    dVC(0+)

    = 0

    dt

    dVa dVC d2i

    = + M (b)

    dt dt dt2

    At t = 0+

    dVa(0+) dVC(0+) d2i(0+)

    = + M (b)

    dt dt dt2

    Putting corresponding values we get

    dVa(0+) V M

    =

    dt L C

    Differentiating (b) with respect to t

    d2Va d

    2VC d

    3i

    = + M

    dt2

    dt2

    dt3

    Differentiating (c) with respect to td

    3i(0+) 1 di(0+) d

    2V(0+)

    L + = (c)

    dt3

    C dt dt2

    d2V(0+)

    =?

    dt2

    V

    d(V(t)) = cos (t/(MC)1/2

    )

    (MC)1/2

    -V

    d2(V(t)) = sin (t/(MC)

    1/2)

    (MC)

    -V

    d2(V(0+)) = sin (0+/(MC)

    1/2)

    (MC)

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    -V

    d2(V(0+)) = sin (0)

    (MC)

    -V

    d2(V(0+)) = (0)

    (MC)

    d2V(0+)

    = 0

    dt2

    d3i(0+)

    = 0 Amp/sec3

    dt3

    Differentiating (c) with respect to t(VCV(t)) d

    2VC

    + C = 0 (c)L dt

    2

    At t = 0+

    (VC(0+)V(0+)) d2VC(0+)

    + C = 0 (c)L dt

    2

    Putting corresponding values

    d2VC(0+)

    = 0 V/sec2

    dt2

    d2Va d

    2VC d

    3i

    = + M

    dt2

    dt2

    dt3

    At t = 0+

    d2Va(0+) d

    2VC(0+) d

    3i(0+)

    = + M

    dt2

    dt2

    dt3

    d2Va(0+)

    = 0 Volt/sec2

    dt2

    Q#5.25: In the network of the figure, the switch K is opened at t = 0 after the

    network has attained a steady state with the switch closed. (a) Find an expression

    for the voltage across the switch at t = 0+. (b) If the parameters are adjusted such

    that i(0+) = 1 and di/dt (0+) = -1, what is the value of the derivative of the voltage

    across the switch, dVK/dt (0+) ?

    CIRCUIT DIAGRAM:

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    + VK -

    R2

    R1 C

    V

    i

    L

    Initial conditions:

    i(0+) = 1

    di(0+)

    = -1

    dt

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    Equivalent network after switching:

    VK

    sc

    V

    iL =

    R2

    V

    iL(0-) = iL(0+) =

    R2

    At t = 0+

    VK(0+) = VR1(0+)

    VR1(0+) = iL(0+)(R1)

    Putting corresponding value we get

    V

    VR1(0+) = R1R2

    For t 01

    VK= iR1 + idtC

    Differentiating with respect to t

    dVK di i

    = R1 +dt dt C

    At t = 0+dVK(0+) di(0+) i(0+)

    = R1 +dt dt C

    Putting corresponding value we get

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    After simplification:

    R2

    Va(0+) = V

    Equivalent network at t =

    VC1() = V.Q#5.27: In the network of the figure, the switch Kis closed at t = 0. At t = 0-, allcapacitor voltages and inductor currents are zero. Three node to datum voltages are

    identified as V1, V2, and V3.(a)Find V1 and dV1/dt at t = 0+.(b)Find V2 and dV2/dt at t = 0+.(c)Find V3 and dV3/dt at t = 0+.

    CIRCUIT DIAGRAM:

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    V1 V3

    V2

    Using KCL at node V1

    For t 0(V1V(t)) dV1 (V1V2) 1

    + C1 + + (V1V3)dt = 0 (i)R1 dt R2 L

    Using KCL at node V2

    (V2V1) dV2 1

    + C2 + V2dt = 0 (ii)R2 dt L2

    Using KCL at node V3

    1 dV3(V3V1)dt + C3 = 0 (iii)L1 dt

    At t = 0+, capacitor C1 becomes short circuit as a result of which

    V1(0+) 0 V

    V2(0+) 0 V

    V3(0+) 0 VAt t = 0+

    1 dV3(V3V1)dt + C3 = 0 (iii)L1 dt

    dV3(0+)

    +

    -

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    iL1(0+) + C3 = 0

    dt

    After simplification we get

    dV3(0+)

    = 0 Volt/secdt

    Here

    iL1(0-) = iL1(0+) = 0 A

    At t = 0+

    (V2(0+)V1(0+)) dV2(0+)+ C2 + iL2(0+) = 0 (ii)

    R2 dt

    iL2(0-) = iL2(0+) = 0 A

    Putting corresponding values we get

    dV2(0+)

    = 0

    dt

    At t = 0+ eq. (i) reveals

    (V1(0+)V(0+)) dV1(0+) (V1(0+)V2(0+)) 1

    + C1 + + (V1(0+)V3(0+))dt = 0R1 dt R2 L

    Putting corresponding values we get

    dV1(0+)

    = 0 Volt/sec.

    dt

    Q#5.28: In the network of the figure, a steady state is reached, and at t = 0, the

    switch K is opened.

    (a)Find the voltage across the switch, VKat t = 0+.(b)Find dVK/dt at t = 0+.

    CIRCUIT DIAGRAM:

    iL3(0+)

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    VK+ -

    Equivalent network before switching

    R1 R2

    i(0-)

    R3

    At t = 0-

    VC1(0-) = i(o-) R2

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    V

    i(0-) =

    R1 + R2 + R3

    VR2

    VC1(0-) =R1 + R2 + R3

    VC1(0-) = i(o-) R2

    VR3

    VC1(0-) =

    R1 + R2 + R3VC2(0-) = VVR1(0-)

    Equivalent network after switching

    V1 K

    At t 0At node K, according to KCL

    d dVk VKV1

    C1 (VKV1) + C3 + = 0dt dt R2

    After simplification we get

    dV1 1 dVK (VKV1)

    = (C3 + C1) + (i)

    dt C1 dt R2

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    At node V1 according to KCL(V1 - V) dV1 d(V1 - VK) (V1 - VK)

    + C2 + C1 + = 0

    R1 dt dt R2After simplification we get

    dV1 1 dVK (VKV1) (VV1)

    = C1 + + (ii)

    dt (C1 + C2) dt R2 R1

    Equating (i) & (ii) we get dVK/dt at t = 0+.

    Hint:

    V1(0+) = VC2(0-) = VVR1(0-)

    Here

    VR1

    VR1(0-) =

    R1 + R2 + R3

    VR2 + VR3

    VVR1(0-) =

    R1 + R2 + R3

    Q#5.29: In the network of the accompanying figure, a steady state is reached with

    the switch K closed and with i = I0, a constant. At t = 0, switch K, is opened. Find:

    (a)V2(0-) =?(b)V2(0+) =?(c) (dV2/dt)(0+).

    CIRCUIT DIAGRAM:1 2

    R2 +

    V2

    C

    R1 R3 L

    -

    Equivalent network at t =0-

    I0

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    After simplification we get

    R1 R2

    Now

    According to current divider rule:R1I0

    iR2 =

    R1 + R2We know V2(0-) = VL(0-) = 0

    Equivalent network at t =0+

    I0

    I0

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    R1

    I0R1 VC(0+)

    iL(0+)

    After simplification

    I0

    + -

    +

    -

    + -

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    V2(0+)

    VC(0+)

    I0R1

    At node V1

    V1 d(V1V2)

    + C = I0 (i)

    R1 dt

    At node V2

    V2 d(V2V1) 1+ C + V2dt (ii)

    R3 dt L

    From eq. (ii)

    d(V1V2) V2 1

    C = + V2dtdt R3 L

    Substituting the value of Cd(V1V2)/dt in (i) we get

    V2 1 V1+ V2dt + = I0

    R3 L R1

    At t = 0+

    V2(0+) 1 V1(0+)

    + V2(0+)dt + = I0 (iii)R3 L R1

    +

    -

    + -

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    Putting corresponding values we get

    I0R1R2

    V1(0+) =

    R1 + R2

    Differentiating eq. (iii) with respect to t and from here putting the value of

    dV1(0+)/dt in eq. (i) we get dV2(0+)/dt.Hint:

    In eq. (i)

    V1(0+) I0R1R2

    =

    R1 (R1 + R2)R1

    dV2(0+) -I0R1R3

    =

    dt C(R1 + R2)(R1 + R3)

    dV1(0+) -R1 dV2(0+)=

    dt R3 dt

    THE END.