In the circuit shown, two resistors $R_{1} a n d R_{2}$ have been connected in series to an ideal cell. When a voltmeter is connected across $R_{1}$ reading is $V_{1} = 4.0$ volt and when the same voltmeter is connected across $R_{2}$ its reading is $V_{2} = 6.0 v o l t$ . The reading of the voltmeter when it is connected across the cell is $V_{3} = 12.0 v o l t$ . The actual voltage across $R_{1}$ in the circuit (in volt) is:

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Detailed Solution

The reading of voltmeter is absolutely correct when it is connected across the battery. It means the Emf of the cell is = 12 volts. Let the resistance of the voltmeter be R. The current I is

$I = \frac{12}{\frac{R_{1} R}{R_{1} + R} + R_{2}}$

$∴$ Voltmeter reading = voltage drop across

$R_{1} (o r R) \Rightarrow 4 = \frac{12}{\frac{R_{1} R}{R_{1} + R} + R_{2}} . \frac{R_{1} R}{R_{1} + R} ∴ 1 + \frac{R_{2} (R_{1} + R)}{R_{1} R} = 3 \Rightarrow \frac{R_{2} (R_{1} + R)}{R_{1} R} = 2 \Rightarrow \frac{R_{2}}{R} + \frac{R_{2}}{R_{1}} = 2 \dots \dots . (1)$

Similarly,

$V_{2} = \frac{12}{\frac{R_{2} R}{R_{2} + R} + R_{1}} . \frac{R_{2} R}{R_{2} + R} \Rightarrow 6 = \frac{12}{1 + \frac{R_{1} (R_{2} + R)}{R_{2} R}} \Rightarrow \frac{R_{1} R_{2} + R_{1} R}{R_{2} R} = 1 \Rightarrow \frac{R_{1}}{R} + \frac{R_{1}}{R_{2}} = 1 \dots \dots (2)$

Eliminating R between (1) and (2)
$\frac{R_{1}}{R_{2}} = \frac{2}{3}$
The actual voltage across $R_{1}$ is
$V_{1} = \frac{12. R_{1}}{R_{1} + R_{2}} = \frac{12}{1 + \frac{R_{2}}{R_{1}}} = \frac{12}{1 + \frac{3}{2}}$