Ncert Exercises & Solutions

Two cells of the same emf E but internal resistance $r_{1}$ and $r_{2}$ are connected in series to an external resistor R (figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero?

Hint: The potential difference across the cells depends on the net current in the circuit.

Step 1: Applying Ohm's law,

Effective resistance

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

and effective emf of two cells = E + E =2E, so the electric current is given by;

I = \frac{E + E}{R + r_{1} + r_{2}}

Step 2: The potential difference across the terminals of the first cell and putting it equal to zero.

V_{1} = E - I_{1} r_{1} = E - (\frac{2 E}{r_{1} + r_{2} + R}) r_{1} = 0

or E = \frac{2 {Er}_{1}}{r_{1} + r_{2} + R} \Rightarrow 1 = \frac{2 r_{1}}{r_{1} + r_{2} + R}

r_{1} + r_{2} + R = 2 r_{1} \Rightarrow R = r_{1} - r_{2}

This is the required relation.

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