Question

A battery of emf $\varepsilon$ and internal resistance $r$ is connected to the circuit as shown in the figure. When the key $K_{1}$ is closed and $K_{2}$ is open, the ideal voltmeter shows a reading of $18\, V$. When key $K_{2}$ is closed and $K_{1}$ is open, the voltmeter reading is $24\, V$. When $K_{1}$ and $K_{2}$ are both open, the voltmeter reading is

• A. 8 V
• B. 12 V
• C. 24 V
• D. 36 V

Answer 1

Answer: (D)

Let $\varepsilon, r$ be emf and internal resistance of battery.
When $K_{2}$ is open and $K_{1}$ is closed, then current through $6\, \Omega$ is
$I_{1}=\frac{\varepsilon}{r+6}=\frac{18}{6}=3$ or $\varepsilon=3 r+18$...(i)
When $K_{1}$ is open and $K_{2}$ is closed, then current through $12\, \Omega$ is
$I_{2}=\frac{\varepsilon}{r+12}=\frac{24}{12}=2$ or $\varepsilon=2 r+24$...(ii)
Solving (i) and (ii), we get $\varepsilon=36\, V , r=6\, \Omega$
When $K_{1}$ and $K_{2}$ are both open, the reading of voltmeter will be the emf of the battery $(=36\, V )$