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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.