Question

A battery of internal resistance, $3\Omega$ is connected to $20\Omega$ resistor and potential difference across the resistor is 10 V. If another resistor of $30\Omega$ is connected in series with the first resistor and battery is again connected to the combination calculate the e.m.f. and terminal potential difference across the combination.

Answer 1

Let the emf of the battery be $E$.

Internal resistance of battery $r=3\Omega$.

Voltage across the resistor $20\Omega$ is $10$ volts.

Voltage across resistor $20\Omega$ $V=\frac{R}{R+r}E$

$\therefore$ $10=\frac{20}{20+3}E$

$⟹$ $E=11.5$ volts

Now another resistor $30\Omega$ is connected in series with battery and previous resistor.

Total resistance of the circuit $R_{eq}=30+20+3=53\Omega$

Current flowing through the circuit $I=\frac{E}{R_{eq}}$

$\therefore$ $I=\frac{11.5}{53}=0.22$ A

Thus terminal potential difference across the combination $V^{\prime }=I(20+30)$

$\therefore$ $V^{\prime }=0.22\times 50=11$ volts