5TUTORIAL 2 DC METERBEE 2123

P1

A PMMC instrument with a 300-turn coil has a 0.15 T magnetic flux density in its air gaps. The coil dimensions are D = 1.25 cm and l = 2 cm. Calculate the torque when the coil current is 500 µA.

N= 300, B = 0.15 T , L= 2x10-2m, D= 1.25x10-2m, I = 500x10-6 A T = BINLD= 0.15 T x 500x10-6 A x 300 x 2x10-2m x 1.25x10-2m =5.625 Nm

P2

A PMMC instrument has a 0.12 T magnetic flux density in its air gaps. The coil dimensions are D = 1.5 cm and l = 2.25 cm. Determine the number of coil turns required to give a torque of 4.5 µNm when the coil current is 100 µA.

B = 0.12T , I = 100x10-6 A , L = 0.0225m , D = 0.015m, T = 4.5x10-6 Nm.T = BINLDN = T/BILD=4.5x10-6 Nm /(0.12T x 100x10-6 A x 0.0225m x = 0.015m) = 1111

P3

A PMMC instrument with a 750 coil resistance gives FSD with a 500 µA coil current. Determine the required shunt resistance to convert the instrument into a dc ammeter with an FSD of (a) 50 mA (b) 30 mA.

Rm = 750 , Im(FSD) = 500 A

(a) 50 mA ranges, Vm = Im Rm

= 500A x 750 = 375 x 10-3 V Is = I – Im = 50mA - 500A = 49.5 x 10-3 A

Vm 375 x 10-3 V Rs1 = --------- = ---------------- I – Im 49.5 x 10-3 A

= 7.576

Im

Rm

Rs1

Rs2

I

1

(b) 30 mA ranges, Is = I – Im = 30mA - 500A = 29.5 x 10-3 A

Vm 375 x 10-3 V Rs2 = --------- = ---------------- I – Im 29.5 x 10-3 A = 12.712

P4

A dc ammeter is constructed of a 133.33 resistance in parallel with a PMMC instrument. If the instrument has a 1.2 k coil resistance and 30 µA FSD, determine the measured current at FSD, 0.5 FSD and 0.33 FSD.

Rm = 1.2 k , Im(FSD) = 30 A , Rs = 133.33

a) at FSD , Vm = IsRs = ImRm = 30 A x 1.2 k

Then, Vm 30 A x 1.2 k Is = ------ = ----------------- = 270 A Rs 133.33

Ranges (current sum) I = Im + Is = 30 A + 270 A = 300 A

b) at 0.5 FSD , IsRs =0.5 ImRm = 15 A x 1.2 k

Then, Vm 30 A x 1.2 k Is = ------ = ----------------- = 135 A Rs 133.33

Ranges (sum current) I = Im + Is = 15 A + 135 A = 150 A

c) at 1/3 FSD , IsRs =1/3 ImRm = 10 A x 1.2 k

Then, Vm 10 A x 1.2 k Is = ------ = ----------------- = 90.02 A Rs 133.33

Ranges (sum current) I = Im + Is = 10 A + 90.02 A = 100 A

Im

Rm

Rs

I

Is

2

P5

A dc ammeter consists of an Ayrton shunt in parallel with a PMMC instrument that has a 1.2 k coil resistance and 100 µA FSD. The Ayrton shunt is made up of four 0.1 series-connected resistors. Calculate the ammeter range at each setting of the shunt.

Rm = 1.2 k , Im(FSD) = 100 A , Rs=0.1

Rm

Im

Rs1 Rs2 Rs3 Rs4

b c a d I

(a) at point A , ImRm = (I – Im) (Rs1 + Rs2 + Rs3 + Rs4) Vs = Is (Rs1 + Rs2 + Rs3 + Rs4) = ImRm

100 A x 1.2 k Is = ---------------------- = Rs1 +Rs2 + Rs3+ Rs4

100 A x 1.2 k = ---------------------------- = 0.3A 0.4

sum I = Is + Im = 300mA + 0.1mA = 300.1 mA

(b) at point B , Im(Rm+ Rs1) = (I – Im) (Rs2 + Rs3 + Rs4)

Vs = Is (Rs2 + Rs3 + Rs4) = Im(Rm+ Rs1)

100 A x (1.2 k +0.1) Is = ------------------------------- Rs2 + Rs3+ Rs4

100 A x 1.2 k = ---------------------------- = 0.4A 0.3

sum I = I s + Im = 400mA + 0.1mA = 400.1 mA

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( c) at point C, Im(Rm+ Rs1 + Rs2) = (I – Im) (Rs3 + Rs4)

Vs = Is (Rs3 + Rs4) = Im(Rm+ Rs1 + Rs2)

100 A x (1.2 k +0.2) Is = ------------------------------- Rs3+ Rs4

100 A x 1200.2 = ---------------------------- = 600.1mA 0.2

sum I = I s + Im = 600.1mA + 0.1mA = 600.2 mA

( d) at point D, Im(Rm+ Rs1 + Rs2 + Rs3) = (I – Im) (Rs4)

Vs = Is (Rs4) = Im(Rm+ Rs1 + Rs2 + Rs3)

100 A x (1.2 k +0.3) Is = ------------------------------- Rs4

100 A x 1200.3 = ---------------------------- = 1200.3mA 0.1

sum I = I s + Im = 1200.3mA + 0.1mA = 1200.4 mA

P6

A 12V source supplies 25A to a load. Calculate the load current that would be measured when using an ammeter with a resistance of (a) 0.12 (b) 0.52 (c) 0.002 .

IL = 25A

RL = 12V/25A

= 0.48 E=12V RL

(a) IL Rm=0.12

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IL(Rm + RL) = 12V

IL = 12/(0.12 + 0.48) E=12V RL

= 20A

(b) IL Rm=0.52

IL(Rm + RL) = 12V

IL = 12/(0.52 + 0.48) E=12V RL

= 12A

(c) IL Rm=0.002

IL(Rm + RL) = 12V

IL = 12/(0.002 + 0.48) E=12V RL

= 24.9A

P7

A PMMC instrument with a 900 coil resistance and an FSD of 75 µA is to be used as a dc voltmeter. Calculate the individual multiplier resistance to give an FSD of (a) 100 V (b) 30 V (c) 5 V. Also, determine the voltmeter sensitivity.

Rm = 900 , Im =75 A

Rs Im= 75 A

(a) Im ( Rm + Rs ) = 100V

Rs = 100/75A - 900

5

= 1.33 M

(b) Im ( Rm + Rs ) = 30V

Rs = 30/75A - 900

= 399 k

(c) Im ( Rm + Rs ) = 5V

Rs = 5/75A - 900

= 65.77 k

Sensitivity = Resistance /Voltage range

(i) Sensitivity = 1.33 M/100 V = 13.3 k/V

(ii) Sensitivity = 399 k/30 V = 13.3 k/V

(iii) Sensitivity = 65.77 k/5 V = 13.15 k/V

P8

A PMMC instrument with Rm = 1.3 k and FSD = 500 µA is used in a multirange dc voltmeter. The series-connected multiplier resistors are R1 = 38.7 k, R2 = 40 k and R3 = 40 k. Calculate the three voltage ranges and determine the voltmeter sensitivity.

Rm = 1.3k, Im= 500A

R1 R2 R3 Rm

B C A Im

(a ) at point A, V = Im ( R1 + R2 + R3 + Rm)

= 500A(38.7k + 40k + 40k +1.3k)

= 60 V

(b ) at point B, V = Im ( R2 + R3 + Rm)

= 500A( 40k + 40k + 1.3k)

= 40.65 V

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(c ) at C point, V = Im (R3 + Rm)

= 500A( 40k + 1.3k)

= 20.65 V

P9

Two resistors, R1 = 47 k and R2 = 82 k, are connected in series across a 15 V supply. A voltmeter on a 10 V range is connected to measure the voltage across R2. The voltmeter sensitivity is 10 k/V. Calculate VR2 (a) with the voltmeter connected (b) with the voltmeter disconnected.

R1 = 47k voltmeter resistance = 10 x 10k/V

E=15V Rv=100k

Rv V R2 = 82k

(a) With voltmeter connected,

Rv//R2 = RvR2/(Rv + R2)

= 100k x 82k /182k

= 45.05 k

VR2 = R2/(Rv//R2 + R1) x 15 V

= 82 k/ (45.05 k + 47 k) x 15 V

= 13.36 V

(b) Without voltmeter connected,

Voltage at R2, R1=47k E = 15V R2 82 k ----------- V = -------------- x 15V R1 + R2 47k+82k

= 82/129 x 15V R2 = 82k = 9.53V

P10

7

A 100 k potentiometer and a 33 k resistor are connected in series across a 9 V supply. Calculate the maximum voltage that can be measured across the potentiometer using a voltmeter with (a) a 20 k/V sensitivity and a 15 V range (b) a 100 k/V sensitivity and a 10 V range.

V Rv

R1=100k R2=33 k

E= 9V

(a) Voltmeter sensitivity 20 k/V Range 15 V

Then, Rv = 20 k/V x 15V = 300k

Max voltage can be measured,

Rv//R1 = 100 k x 300 k/ 400 k = 75 k

VR1 = 75 k/(75 k + 33 k) x 9V = (75/108) x 9V = 6.25V

(b) Voltmeter sensitivity100k/V Ranges 10 V

Then, Rv = 100 k/V x 10V = 1000k Rv//R1 = 1000 k x 100 k / 1100 k = 90.9 k

VR1 = 90.9 k /(90.9 k + 33 k) 9V = (90.9/123.9) x 9V = 6.6V

P11

A series ohmmeter is made up of the following components: supply voltage Eb = 3 V, series resistor R1 = 30 k, meter shunt resistor R2 = 50 , meter FSD = 50 µA, and meter resistance Rm = 50 . Determine the resistance measured at (a) 0.25 FSD (b) 0.5 FSD (c) 0.75 FSD (d) FSD.

Rx

R1=30k

Ib

Eb=3V R2=50 Rm=50

I2 Im=50A

(a). At 0.25 FSD and 3V supply, Im=50/4A =12.5A Rx + R1 = 3V/ Ib

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Rx = 3V/ Ib - R1 But, Ib = Im + I2

And I2 = ImRm/R2

=12.5Ax 50/50 =12.5A Then Ib = 12.5A+12.5A = 25A

So, Rx = 3V/25A-30k = 90k

(b) At 0.5 FSD and 3V supply, Im=50/2A =25A Rx + R1 = 3V/ Ib

Rx = 3V/ Ib - R1 But, Ib = Im + I2

And I2 = ImRm/R2

=25Ax 50/50 =25A Then Ib = 25A+25A = 50A

So, Rx = 3V/50A-30k = 30k

(c) At 0.75FSD and 3V supply, Im=(3x50/4)A =37.5A Rx + R1 = 3V/ Ib

Rx = 3V/ Ib - R1 But, Ib = Im + I2

And I2 = ImRm/R2

=37.5Ax 50/50 =37.5A Then Ib = 37.5A+37.5A = 75A

So, Rx = 3V/75A-30k = 10k

(d) At FSD and 3V supply, Im=50A Rx + R1 = 3V/ Ib

Rx = 3V/ Ib - R1

But, Ib = Im + I2

And I2 = ImRm/R2

=50Ax 50/50 =50A Then Ib = 50A+50A = 100A

So, Rx = 3V/100A-30k = 0

P12

A series ohmmeter that has a standard internal resistance of R1 = 50 k uses a meter with FSD = 75 µA and Rm = 100 . The meter shunt resistance is R2 = 300 and the battery voltage Eb = 5 V. Determine the resistance measured at (a) 0.25 FSD (b) 0.5 FSD (c) 0.75 FSD (d) FSD.

Rx

R1=50 k,

Ib

9

Eb=5V R2=300 Rm=100 I2 Im=75A

(a) At 0.25 FSD and 5V supply, Im=75/4A =18.75A Rx + R1 + R2//Rm= 5V/ Ib

Rx = 5V/ Ib - R1- R2//Rm But, Ib = Im + I2

I2 = ImRm/R2

=18.75A x100/300 =6.25A Then, Ib = 18.75A +6.25A = 25A

So, Rx = 5V/25A-50k - R2//Rm

= 5V/25A-50k =150 k

(b) At 0.5 FSD and 5V supply, Im=75/2A =37.5A Rx + R1 = 5V/ Ib

But, Ib = Im + I2

I2 = ImRm/R2

=37.5A x100/300 =12.5A Then, Ib = 37.5A +12.5A = 50A

So, Rx = 5V/50A-50k = 50k

(c) At 0.75 FSD and 5V supply, Im=(3x75/4)A =56.25A Rx + R1 = 5V/ Ib

But, Ib = Im + I2

I2 = ImRm/R2

=56.25A x100/300 =18.75A Then, Ib = 56.25A +18.75A = 75A

So, Rx = 5V/75A-50k = 16.67k

(d) At FSD and 5V supply, Im=75A Rx + R1 = 5V/ Ib

But, Ib = Im + I2

I2 = ImRm/R2

=75A x100/300 =25A Then, Ib = 75A +25A = 100A

So, Rx = 5V/100A-50k = 0

P13

For the ohmmeter circuit in Problem 1, determine the new resistance to which R2 must be adjusted when Eb falls to 2.5 V. Also, determine the new resistances measured at (a) 0.5 FSD (b) 0.75 FSD.

Rx

R1=30k

10

Ib

Eb=2.5V R2 Rm=50

I2 Im=50A

When Rx=0, Ib = 2.5V/(30k+R2//Rm) But R2//Rm is very small compared to R1 therefore can be neglected. Then, Ib = 2.5V/30k =83.33A But I2 = Ib - Im = 83.33A - 50A =33.33A So, R2 = ImRm/I2 = 50Ax50/33.33A = 75

(a) At 0.5 FSD and 2.5V supply, Im=50/2A =25A Rx + R1 = 2.5V/ Ib

Rx = 2.5V/ Ib - R1

But, Ib = Im + I2

I2 = ImRm/R2

=25A x50/75 =16.67A Then Ib = 25A +16.67A = 41.6A

So, Rx = 2.5V/41.6A-30k = 29.995k

30k

(b) At 0.75 FSD and 2.5V supply, Im=0.75x50A =37.5A Rx + R1 = 2.5V/ Ib

Rx = 2.5V/ Ib - R1

But, Ib = Im + I2

I2 = ImRm/R2

=37.5A x50/75 =25A Then Ib = 37.5A +25A = 62.5A

So, Rx = 2.5V/62.5A-30k = 10 k

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