# Tutorial 2

### Transcript of Tutorial 2

Tutorial 2

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Problem 1

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.

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SolutionGiven:N= 300, B = 0.15 TL = 2 x 10-2 mD= 1.25 x 10-2 mI = 500x10-6 ATorque:τ = B x A x I x N = B x l x D x I x N

= 0.15 T x 2x10-2m x 1.25x10-2m x 500 x 10-6 A x 300 =5.625 μNm

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Problem 2

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.

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Solution

B = 0.12 T I = 100x10-6 A l = 0.0225 mD = 0.015 mτ = 4.5 x 10-6 Nmτ = BAIN = BlDINN = τ /BlDI = 4.5x10-6 Nm /(0.12T x 0.0225 m x

0.015 m x 100 x 10-6 A) = 1111

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Problem 3

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.

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Solution

Rm = 750Ω , Ifsd = 500 μA(a) 50 mA ranges,

Vm = ImRm = IfsdRm = 500μA x 750Ω= 375 x 10-3 V

Ish1= I – Im = I – Ifsd = 50mA - 500μA = 49.5 x 10-3 A

Im

Rm

Rsh1

Rsh2

IΩ=

××

== −

−

576.7105.4910375

3

3

11 A

VIVR

sh

msh

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Solution

(b) 30 mA ranges, Ish2= I – Im = I – Ifsd = 30mA - 500μA

= 29.5 x 10-3 A

Ω=××

== −

−

712.12105.2910375

3

3

22 A

VIVR

sh

msh

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Problem 4

A dc ammeter is constructed of a 133.33 Ωresistance 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.

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Solution

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Problem 5

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.

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Solution

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Problem 6

A 12 V source supplies 25 A 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

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Solution

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Problem 7

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.

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Solution

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Problem 8

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.

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Solution

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Problem 9

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.

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Solution

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Problem 10

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.

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Solution

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Problem 11

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.

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Solution

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Problem 12

A series ohmmeter that has a standard internal resistance of R1 = 50 kΩ uses a meter with FSD = 75 µA ad 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.

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Solution

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Problem 13

For the ohmmeter circuit in Problem 12, determine the new resistance to which R2must be adjusted when Eb falls to 2.5 V. Also, determine the new resistances measured at

(a) 0.5 FSD (b) 0.75 FSD.

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Solution

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