BEE2123 TUT2_DCMeter Solution

5TUTORIAL 2 DC METER BEE 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 µ 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 = BINLD N = 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 Ω 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 = 750Ω , Im(FSD) = 500 µ A Rm Im Rs1 Rs2

I (a) 50 mA ranges,

Vm = Im Rm = 500µ A x 750Ω = 375 x 10-3 V Is = I – Im = 50mA - 500µ A = 49.5 x 10-3 A Rs1

Vm 375 x 10-3 V = --------- = ---------------I – Im 49.5 x 10-3 A = 7.576 Ω

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(b) 30 mA ranges, Is = I – Im = 30mA - 500µ A = 29.5 x 10-3 A Vm 375 x 10-3 V Rs2 = --------- = ---------------I – Im 29.5 x 10-3 A = 12.712 Ω P4 A dc ammeter is constructed of a 133.33 Ω resistance in parallel with a PMMC instrument. If the instrument has a 1.2 kΩ coil resistance and 30 µA FSD, determine the measured current at FSD, 0.5 FSD and 0.33 FSD. Rm = 1.2 kΩ , Im(FSD) = 30 µ A , Rs = 133.33 Ω Rm Im Rs Is I a) at FSD ,

Vm = IsRs = ImRm = 30 µ A x 1.2 kΩ

Then, Is

Vm 30 µ A x 1.2 kΩ = ------ = ----------------- = 270 µ A Rs 133.33 Ω

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

b) at 0.5 FSD ,

IsRs =0.5 ImRm = 15 µ A x 1.2 kΩ

Then, Is

Vm 30 µ A x 1.2 kΩ = ------ = ----------------- = 135 µ A Rs 133.33 Ω

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

c) at 1/3 FSD ,

= 150 µ A

IsRs =1/3 ImRm = 10 µ A x 1.2 kΩ

Then, Is

Vm 10 µ A x 1.2 kΩ = ------ = ----------------- = 90.02 µ A Rs 133.33 Ω

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Ranges (sum current) I = Im + Is = 10 µ A + 90.02 µ A = 100 µ A P5 A dc ammeter consists of an Ayrton shunt in parallel with a PMMC instrument that has a 1.2 kΩ coil resistance and 100 µA FSD. The Ayrton shunt is made up of four 0.1 Ω series-connected resistors. Calculate the ammeter range at each setting of the shunt. Rm = 1.2 kΩ , Im(FSD) = 100 µ A , Rs=0.1Ω 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 µ A x 1.2 kΩ Is = ---------------------- = Rs1 +Rs2 + Rs3+ Rs4 100 µ A x 1.2 kΩ = ---------------------------- = 0.3A 0.4Ω 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 µ A x (1.2 kΩ +0.1Ω ) Is = ------------------------------Rs2 + Rs3+ Rs4 100 µ A x 1.2 kΩ = ---------------------------- = 0.4A 0.3Ω

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sum I = I s + Im = 400mA + 0.1mA = 400.1 mA ( c) at point C,

Im(Rm+ Rs1 + Rs2) = (I – Im) (Rs3 + Rs4)

Vs = Is (Rs3 + Rs4) = Im(Rm+ Rs1 + Rs2) 100 µ A x (1.2 kΩ +0.2Ω ) Is = ------------------------------Rs3+ Rs4 100 µ A x 1200.2 Ω = ---------------------------- = 600.1mA 0.2Ω 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 µ A x (1.2 kΩ +0.3Ω ) Is = ------------------------------Rs4 100 µ A x 1200.3 Ω = ---------------------------- = 1200.3mA 0.1Ω 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 Ω (b) 0.52 Ω (c) 0.002 Ω . IL = 25A RL = 12V/25A E=12V

RL

= 0.48Ω

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(a)

Rm=0.12Ω

IL

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

RL = 20A

(b)

Rm=0.52Ω

IL

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

RL = 12A

(c)

Rm=0.002Ω

IL

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

RL = 24.9A

P7 A PMMC instrument with a 900 Ω 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 = 900Ω , Im =75 µ A Rs

Im= 75 µ A

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

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Rs = 100/75µ A - 900Ω = 1.33 MΩ (b) Im ( Rm + Rs ) = 30V Rs = 30/75µ A - 900Ω = 399 kΩ (c) Im ( Rm + Rs ) = 5V Rs = 5/75µ A - 900Ω = 65.77 kΩ Sensitivity = Resistance /Voltage range (i)

Sensitivity = 1.33 MΩ /100 V = 13.3 kΩ /V

(ii) Sensitivity = 399 kΩ /30 V

= 13.3 kΩ /V

(iii) Sensitivity = 65.77 kΩ /5 V

= 13.15 kΩ /V

P8 A PMMC instrument with Rm = 1.3 kΩ and FSD = 500 µA is used in a multirange dc voltmeter. The series-connected multiplier resistors are R1 = 38.7 kΩ , R2 = 40 kΩ and R3 = 40 kΩ . Calculate the three voltage ranges and determine the voltmeter sensitivity. Rm = 1.3kΩ ,

Im= 500µ A

R1

R2

B

R3

Rm

C

A Im

(a ) at point A,

V = Im ( R1 + R2 + R3 + Rm) = 500µ A(38.7kΩ + 40kΩ + 40kΩ +1.3kΩ ) = 60 V

(b ) at point B,

V = Im ( R2 + R3 + Rm) = 500µ A( 40kΩ + 40kΩ + 1.3kΩ )

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= 40.65 V (c ) at C point,

V = Im (R3 + Rm) = 500µ A( 40kΩ + 1.3kΩ ) = 20.65 V

P9 Two resistors, R1 = 47 kΩ and R2 = 82 kΩ , 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 kΩ /V. Calculate VR2 (a) with the voltmeter connected (b) with the voltmeter disconnected.

R1 = 47kΩ

voltmeter resistance = 10 x 10kΩ /V Rv=100kΩ

E=15V

Rv

R2 = 82kΩ

V

(a) With voltmeter connected, Rv//R2

= RvR2/(Rv + R2) = 100kΩ x 82kΩ /182kΩ = 45.05 kΩ

VR2

= R2/(Rv//R2 + R1) x 15 V = 82 kΩ / (45.05 kΩ + 47 kΩ ) x 15 V = 13.36 V

(b) Without voltmeter connected,

R1=47kΩ E = 15V

R2 = 82kΩ

Voltage at R2, R2 82 kΩ ----------- V = -------------- x 15V R1 + R2 47kΩ +82kΩ = 82/129 x 15V = 9.53V

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P10 A 100 kΩ potentiometer and a 33 kΩ 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 kΩ /V sensitivity and a 15 V range (b) a 100 kΩ /V sensitivity and a 10 V range.

V

Rv

R1=100kΩ

R2=33 kΩ

E= 9V (a)

Voltmeter sensitivity 20 kΩ /V Range 15 V Then, Rv = 20 kΩ /V x 15V = 300kΩ Max voltage can be measured, Rv//R1 = 100 kΩ x 300 kΩ / 400 kΩ = 75 kΩ VR1 = 75 kΩ /(75 kΩ + 33 kΩ ) x 9V = (75/108) x 9V = 6.25V

(b)

Voltmeter sensitivity100kΩ /V Ranges 10 V Then, Rv = 100 kΩ /V x 10V = 1000kΩ Rv//R1 = 1000 kΩ x 100 kΩ / 1100 kΩ = 90.9 kΩ VR1 = 90.9 kΩ /(90.9 kΩ + 33 kΩ ) 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 kΩ , meter shunt resistor R2 = 50 Ω , meter FSD = 50 µA, and meter resistance Rm = 50 Ω . Determine the resistance measured at (a) 0.25 FSD (b) 0.5 FSD (c) 0.75 FSD (d) FSD. Rx R1=30kΩ Ib Eb=3V

R2=50Ω

Rm=50Ω

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I2

Im=50µ A

(a). At 0.25 FSD and 3V supply, Im=50/4µ A =12.5µ A Rx + R1 = 3V/ Ib Rx = 3V/ Ib - R1 But, Ib = Im + I2 And I2 = ImRm/R2 =12.5µ Ax 50Ω /50Ω =12.5µ A Then Ib = 12.5µ A+12.5µ A = 25µ A So, Rx = 3V/25µ A-30kΩ = 90kΩ (b) At 0.5 FSD and 3V supply, Im=50/2µ A =25µ A Rx + R1 = 3V/ Ib Rx = 3V/ Ib - R1 But, Ib = Im + I2 And I2 = ImRm/R2 =25µ Ax 50Ω /50Ω =25µ A Then Ib = 25µ A+25µ A = 50µ A So, Rx = 3V/50µ A-30kΩ = 30kΩ (c) At 0.75FSD and 3V supply, Im=(3x50/4)µ A =37.5µ A Rx + R1 = 3V/ Ib Rx = 3V/ Ib - R1 But, Ib = Im + I2 And I2 = ImRm/R2 =37.5µ Ax 50Ω /50Ω =37.5µ A Then Ib = 37.5µ A+37.5µ A = 75µ A So, Rx = 3V/75µ A-30kΩ = 10kΩ (d) At FSD and 3V supply, Im=50µ A Rx + R1 = 3V/ Ib Rx = 3V/ Ib - R1 But, Ib = Im + I2 And

I2 = ImRm/R2 =50µ Ax 50Ω /50Ω =50µ A Then Ib = 50µ A+50µ A = 100µ A So, Rx = 3V/100µ A-30kΩ = 0Ω P12

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A series ohmmeter that has a standard internal resistance of R1 = 50 kΩ uses a meter with FSD = 75 µA and Rm = 100 Ω . The meter shunt resistance is R2 = 300 Ω 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

Ib Eb=5V

R1=50 kΩ ,

R2=300Ω I2

Rm=100Ω Im=75µ A

(a) At 0.25 FSD and 5V supply, Im=75/4µ A =18.75µ A Rx + R1 + R2//Rm= 5V/ Ib Rx = 5V/ Ib - R1- R2//Rm But, Ib = Im + I2 I2 = ImRm/R2 =18.75µ A x100Ω /300Ω =6.25µ A Then, Ib = 18.75µ A +6.25µ A = 25µ A So, Rx = 5V/25µ A-50kΩ - R2//Rm = 5V/25µ A-50kΩ =150 kΩ (b) At 0.5 FSD and 5V supply, Im=75/2µ A =37.5µ A Rx + R1 = 5V/ Ib But, Ib = Im + I2 I2 = ImRm/R2 =37.5µ A x100Ω /300Ω =12.5µ A Then, Ib = 37.5µ A +12.5µ A = 50µ A So, Rx = 5V/50µ A-50kΩ = 50kΩ (c) At 0.75 FSD and 5V supply, Im=(3x75/4)µ A =56.25µ A Rx + R1 = 5V/ Ib But, Ib = Im + I2 I2 = ImRm/R2 =56.25µ A x100Ω /300Ω =18.75µ A Then, Ib = 56.25µ A +18.75µ A = 75µ A So, Rx = 5V/75µ A-50kΩ = 16.67kΩ (d) At FSD and 5V supply, Im=75µ A Rx + R1 = 5V/ Ib But, Ib = Im + I2 I2 = ImRm/R2 =75µ A x100Ω /300Ω =25µ A Then, Ib = 75µ A +25µ A = 100µ A

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So, Rx = 5V/100µ A-50kΩ = 0Ω 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=30kΩ Ib Eb=2.5V

Rm=50Ω

R2 I2

Im=50µ A

When Rx=0, Ib = 2.5V/(30kΩ +R2//Rm) But R2//Rm is very small compared to R1 therefore can be neglected. Then, Ib = 2.5V/30kΩ =83.33µ A But I2 = Ib - Im = 83.33µ A - 50µ A =33.33µ A So, R2 = ImRm/I2 = 50µ Ax50Ω /33.33µ A = 75Ω (a) At 0.5 FSD and 2.5V supply, Im=50/2µ A =25µ A Rx + R1 = 2.5V/ Ib Rx = 2.5V/ Ib - R1 But, Ib = Im + I2 I2 = ImRm/R2 =25µ A x50Ω /75Ω =16.67µ A Then Ib = 25µ A +16.67µ A = 41.6µ A

≈

30kΩ

So, Rx = 2.5V/41.6µ A-30kΩ = 29.995kΩ

(b) At 0.75 FSD and 2.5V supply, Im=0.75x50µ A =37.5µ A Rx + R1 = 2.5V/ Ib Rx = 2.5V/ Ib - R1 But, Ib = Im + I2 I2 = ImRm/R2 =37.5µ A x50Ω /75Ω =25µ A Then Ib = 37.5µ A +25µ A = 62.5µ A So, Rx = 2.5V/62.5µ A-30kΩ = 10 kΩ

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