A battery of emf 10V and internal resistance 3 $Ω$ is connected to a resistor. If the current in the circuit is 0.5A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?

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Detailed Solution

$E = 10 V, r = 3 Ω, I = 0.5 A$
Let ‘R’ be the resistance of the resistor

$I = \frac{E}{R + r} R = \frac{E}{I} - r = \frac{10}{0 .5} - 3 = 17 Ω$
Terminal voltage of battery
$V = E - Ir = 10 - 0 .5 \times 3 = 8 .5 V$

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