Calculate the steady state current in the 2 Ω resistor shown in the figure below. The internal resistance of the battery is negligible and the capacitance of the capacitor is 0.2 μF.

Solution

The resistances 2Ω and 3Ω are connected in parallel.
Therefore, equivalent resistance of 2Ω and 3 Ω in parallel will be

R_{p} = \frac{2 \times 3}{2 + 3} = \frac{6}{5} = 1.2 Ω

Since, capacitor provides infinite resistance to direct current, hence no current flows in capacitor arm.

Therefore, total resistance of current carrying arms
R = 1.2 + 2.8 = 4 Ω

Hence current from battery will be
$I = \frac{6}{4} = 1.5 A$

∴ Potential difference across A and B = IR_p= 1.5 x 1.2 = 1.80 V

Current through 2 Ω resistor

$= \frac{potential difference}{resistance}$
$= \frac{1.8}{2} = 0.9 A .$

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Current Electricity

Calculate the steady state current in the 2 Ω resistor shown in the figure below. The internal resistance of the battery is negligible and the capacitance of the capacitor is 0.2 μF.

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