Page 252 - Basic Electrical Engineering

V = IR drops across 4 kΩ and 6 kΩ resistors
PQ

Now, when the switch is opened, at t = 0, v = 10 V. The capacitor will be
c
getting discharged through the resistors 10 kΩ, 4 kΩ, and 6 kΩ in the loop
RPLMQSR. The time constant of the circuit, τ = RC.

−6
3
τ = RC = (10 + 4 + 6) × 10 × 2.5 × 10 = 20 × 103 × 2.5 × 10 −6
= 0.05 seconds

Let the capacitor voltage at t = 0.05 sec be and discharging current at t =
′
0.05 be i .

Initial current at t = 0,

Current after 0.05 sec,

−1
i′ = I e = 1 × 0.368 mA = 0.368mA
0

2.11 REVIEW QUESTIONS

A. Short Answer Type Questions

1. Define Ohm’s law and state if there are any conditions.
2. Explain the concept of voltage and current source transformation with an example.
3. Give the concept of current, voltage, and resistance.
4. State the factors on which resistance of a wire depends. What is meant by resistivity of a
conducting material?
5. Explain why silver is more conducting than copper.