In a network as shown in the figure the potential difference across the resistance 2R (the cell has an emf of E and has no internal resistance) :-

Question

Verified by Toppr

Answer 1

Total resistance of circuit is $R_{t} = R + \frac{2 R . 4 R}{2 R + 4 R} = \frac{7 R}{3}$

Current in the circuit is $I = \frac{E}{R _{t}} = 3 E / 7 R$

Potential across R is $V_{1} = I R = 3 E / 7$

As R and (2R, 4R) are in series so potential across (2R,4R) is $V_{2} = E - V_{1} = E - 3 E / 7 = 4 E / 7$ . This is also potential across 2R because 2R and 4R are in parallel.

Answer 2

Total resistance of circuit is

R_{t} = R + \frac{2 R . 4 R}{2 R + 4 R} = \frac{7 R}{3}

Current in the circuit is

I = \frac{E}{R _{t}} = 3 E / 7 R

Potential across R is

V_{1} = I R = 3 E / 7

As R and (2R, 4R) are in series so potential across (2R,4R) is

V_{2} = E - V_{1} = E - 3 E / 7 = 4 E / 7

. This is also potential across 2R because 2R and 4R are in parallel.