Question

Consider a circuit having resistance 10 kΩ, excited by voltage 5 V and an ideal switch S. If the switch is repeatedly closed for 2 ms and opened for 2 ms, the average value of i(t) is ____________

(a) 0.25 mA

(b) 0.35 mA

(c) 0.125 mA

(d) 1 mA

Answer 1

Correct option is (c) 0.125 mA

The explanation: Since i = \(\frac{5}{10 × 2 X 10^{-3}}\) = 0.25 × 10^-3 = 0.25 mA.

As the switch is repeatedly close, then i (t) will be a square wave.

So average value of electric current is \((\frac{0.25}{2})\) = 0.125 mA.