First slide

Ohm's law

Question

The potential difference in open circuit for a cell is 2.2 volts. When a 4 ohm resistor is connected between its two electrodes the potential difference becomes 2 volts. The internal resistance of the cell will be

Moderate

A

1 ohm
B

0.2 ohm
C

2.5 ohm
D

0.4 ohm

Solution

The potential difference in open circuit for the cell is 2.2 volts, it means the emf of the battery should be 2.2 volts.
Now a 4 ohm resistor is connected between its two electrodes, then the potential difference across the battery becomes 2 volts.
$\begin{array}{l} V = ε - ir \Rightarrow 2 = 2 .2 - ir \\ \Rightarrow ir = 0 .2 V . ... (i) \\ But i = \frac{ε}{r + R} = \frac{2 .2}{r + 4} . .... (ii) \end{array}$
Putting the value of i in (i), we get r = 0.4 ohm.

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