JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Current Electricity

Two cells of emf 2E and E with internal resistance

r_{1}

and

r_{2}

respectively are connected in series to an external resistor R (see figure). The value of R , at which the potential difference across the terminals of the first cell becomes zero is

(A) $\frac{r _{1}}{2} - r_{2}$ ✓ Correct
(B) $\frac{r _{1}}{2} + r_{2}$
(C) $r_{1} + r_{2}$
(D) $r_{1} - r_{2}$

Answer: (A)

Solution

For the potential difference across the first cell (emf = 2E, internal resistance =

r_{1}

) to be zero:

V_{1} = 2 E - I r_{1} = 0

This gives:

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

The total emf in the circuit is

2 E + E = 3 E

, and total resistance is

r_{1} + r_{2} + R

.

Current:

I = \frac{3 E}{r _{1} + r _{2} + R}

Equating the two expressions for current:

\frac{2 E}{r _{1}} = \frac{3 E}{r _{1} + r _{2} + R}

2 (r_{1} + r_{2} + R) = 3 r_{1}

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

R = \frac{r _{1}}{2} - r_{2}

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Current Electricity

Solution

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