Answer 1
Under steady state conditions, the capacitor is an open circuit and so the voltage across it is equal to the voltage between the terminals A and B. No current flows through the rightmost 1K resistor and therefore there is no drop across it.

The effective resistance from the battery's point of view is (1K + 1K) in parallel with 2K and then that combination is in series with 1K. The total effective resistance is 2K || 2K + 1K = 2K.

The current I drawn from the battery is 10V / 2K = 5mA. The current I' that flows through the center resistor is I × 2K/(2K + 2K) = I/2 = 2.5mA. The voltage drop across the resistor between A and B = 2.5 mA × 1K = 2.5V. The voltage across the 1 µF capacitor is therefore 2.5V.

Contributor: Naveen PN

Published: September-2002