# Electromagnetic Theory Questions and Answers – Resistances and Capacitances

This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Resistance and Capacitance”.

1. Find the resistivity of a material having resistance 20kohm, area 2 units and length of 12m.
a) 6666.6
b) 3333.3
c) 1200
d) 2000

Explanation: The resistance of a material is given by R = ρL/A. To get ρ, put R = 20 x 103, A = 2 and L = 12. We get ρ = 3333.3 units.

2. A resistor value of colour code orange violet orange will be
a) 37 kohm
b) 37 Mohm
c) 48 kohm
d) 48 Mohm

Explanation: Orange refers to number 3. Violet refers to number 7. The third colour code orange refers to 103. Thus the resistor value will be 37 kilo ohm.

3. A infinite resistance is considered as a/an
a) Closed path(short circuit)
b) Open path
c) Not defined

Explanation: When there exists infinite resistance in a path, the current flowing will ideally be zero. This is possible only for an open path/circuit.

4. Find the time constant in a series R-L circuit when the resistance is 4 ohm and the inductance is 2 H.
a) 0.25
b) 0.2
c) 2
d) 0.5

Explanation: The time constant for an R-L series circuit will be τ = L/R. Put R = 4 and L = 2. We get τ = 2/4 = 0.5 second.

5. Find the time constant for a R-C circuit for resistance R = 24 kohm and C = 16 microfarad.
a) 1.5 millisecond
b) 0.6 nanosecond
c) 384 millisecond
d) 384 microsecond

Explanation: The time constant for R-C circuit is τ = RC. Put R = 24 kilo ohm and C = 16 micro farad. We get τ = 24 x 103 x 16 x 10-6 = 0.384 = 384 millisecond.
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6. Find the capacitance when charge is 20 C has a voltage of 1.2V.
a) 32.67
b) 16.67
c) 6.67
d) 12.33

Explanation: Capacitance is related to Q and V as C = Q/V. Put C = 20C and V = 1.2V, we get Q = 20/1.2 = 16.67 farad.

7. Calculate the capacitance of two parallel plates of area 2 units separated by a distance of 0.2m in air(in picofarad)
a) 8.84
b) 88.4
c) 884.1
d) 0.884

Explanation: Capacitance is given by, C = εo A/d. Put A = 2, d = 0.2, εo = 8.854 x 10-12, we get C = 8.841 x 10-11 = 88. 41 pF.

8. Compute the capacitance between two concentric shells of inner radius 2m and the outer radius is infinitely large.
a) 0.111 nF
b) 0.222 nF
c) 4.5 nF
d) 5.4 nF

Explanation: The concentric shell with infinite outer radius is considered to be an isolated sphere. The capacitance C = 4πε/(1/a – 1/b). If b->∞, then C = 4πεa. Put a = 2m, we get C = 4π x 8.854 x 10-12 x 2 = 0.222 nF.

9. The capacitance of a material refers to
a) Ability of the material to store magnetic field
b) Ability of the material to store electromagnetic field
c) Ability of the material to store electric field
d) Potential between two charged plates

Explanation: The capacitance of a material is a measure of the ability of the material to store electric field. It is the ratio of charge stored to the voltage across the parallel plates.

10. A cable of core radius 1.25cm and impregnated paper insulation of thickness 2.13cm and relative permittivity 3.5. Compute the capacitance of the cable/km(in nF)
a) 195.7
b) 179.5
c) 157.9
d) 197.5

Explanation: Capacitance between coaxial cylinders of inner radius 1.25cm and outer radius 1.25 + 2.13 = 3.38cm will be C = 2πεL/ ln(b/a). Put b = 3.38, a = 1.25 and L = 1000m, we get C = 1.957 x 10-7 = 195.7 nF.

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To practice all areas of Electromagnetic Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.

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