A 1μF capacitor is connected in the circuit shown below. The emf of the cell is 3V and internal resistance is 0.5Ω. The resistors R1 and R2 have values 4Ω and 1Ω respectively. The charge on the capacitor in steady state must be

By Shivank Goel|Updated : August 17th, 2022

A 1μF capacitor is connected in the circuit shown below

Given, a 1μF capacitor is connected to the circuit.

Emf of the cell, E = 3V

Internal resistance, r = 0.5Ω

Resistance of the resistors R1 and R2 are 4Ω and 1Ω.

We have to find the charge on the capacitor in a steady state.

In a steady state, no current flows through the capacitor.

The potential difference across capacitor and resistor of resistance R₂ is the same. So, the emf (3V) is shared between r and R₂.

We know, charge on capacitor = CV

Where V is the voltage across the capacitor

Here, V = (R₂/r+R₂)xE

V = 1/(0.5+1) x 3

V = 1/1.5 x 3

V = 2

Now, Q = 1x2

= 2μC

Therefore, the required charge on the capacitor is 2μC.

Summary:

A 1μF capacitor is connected in the circuit shown below. The emf of the cell is 3V and the internal resistance is 0.5Ω. The resistors R1 and R2 have values of 4Ω and 1Ω respectively. The charge on the capacitor in a steady state must be

A 1μF capacitor is connected in the circuit shown below. The emf of the cell is 3V and the internal resistance is 0.5Ω. The resistors R1 and R2 have values of 4Ω and 1Ω respectively. The charge on the capacitor in a steady state must be 2μC.

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