JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Current Electricity

Two cells of same emf but different internal resistances

r_{1}

and

r_{2}

are connected in series with a resistance

R

. The value of resistance

R

, for which the potential difference across second cell is zero, is

Answer: (A)

Solution

Current in the series circuit:

I = \frac{2 E}{R + r _{1} + r _{2}}

. For the potential difference across the second cell to be zero:

E - I r_{2} = 0

, so

I = \frac{E}{r _{2}}

. Equating the two expressions for I:

\frac{E}{r _{2}} = \frac{2 E}{R + r _{1} + r _{2}}

. Simplifying:

R + r_{1} + r_{2} = 2 r_{2}

, giving

R = r_{2} - r_{1}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option