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Calculate potential of a metal plate of charge 28C and capacitance 12 mF.
b Explanation: Potential is given by V = Q/C. Put Q = 28C and C = 12 mF. We get V = 28/12 x 10-3 = 2.333 x 103 ohm.
b
See lessExplanation: Potential is given by V = Q/C. Put Q = 28C and C = 12 mF. We get V =
28/12 x 10-3 = 2.333 x 103 ohm.
Calculate the electric field intensity of a line charge of length 2m and potential 24V.
b Explanation: The electric field intensity is given by the ratio of potential to distance or length. E = V/d = 24/2 = 12 V/m.
b
See lessExplanation: The electric field intensity is given by the ratio of potential to distance or
length. E = V/d = 24/2 = 12 V/m.
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 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
a
See lessExplanation: 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
The capacitance of a material refers to
c 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.
c
See lessExplanation: 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.
Compute the capacitance between two concentric shells of inner radius 2m and the outer radius is infinitely large.
b 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
b
See lessExplanation: 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
Calculate the capacitance of two parallel plates of area 2 units separated by a distance of 0.2m in air(in picofarad)
b 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.
b
See lessExplanation: 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.
Find the capacitance when charge is 20 C has a voltage of 1.2V.
b 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.
b
See lessExplanation: 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.
Find the time constant in a series R-L circuit when the resistance is 4 ohm and the inductance is 2 H.
d 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.
d
See lessExplanation: 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.
A infinite resistance is considered as a/an
b 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.
b
See lessExplanation: When there exists infinite resistance in a path, the current flowing will
ideally be zero. This is possible only for an open path/circuit.
A resistor value of colour code orange violet orange will be
a 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.
a
See lessExplanation: 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.