Calculate the steady-state current in the 2 Ω resistor shown in Fig. The internal resistance of the battery is negligible and capacitance of the condenser is 0.2 μ F.

In the steady state, no current flows through the capacitor C and hence no current passes through the 4 resistor which is in series with the capacitor.

Solution

Resistance 2Ω and 3Ω are grouped in parallel combination.
Therefore, their effective resistance is R₁₂

\frac{1}{R_{12}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}

R_{12} = 1.2 Ω

Resistance R₁₂ is in series with 2.8 Ω.
Their total effective resistance = 1.2 + 2.8 = 4.0 Ω.

Thus, the current through the circuit, I= 6/4 =1.5 A.

Hence, the potential difference across AB = 1.5 x 1.2 = 1.8 V
and,
The current through 2Ω resistor

= \frac{1.8}{2} = 0.9 A .

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