Question

A battery of emf $\epsilon$ 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 $18V$. When key $K_2$ is closed and $K_1$ is open, the voltmeter reading is $24V$. 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 $\epsilon ,r$ be emf and internal resistance of battery.
When $K_2$ is open and $K_1$ is closed, then current through $6\Omega$ is
$I_1=\frac{\epsilon }{r+6}=\frac{18}{6}=3$ or $\epsilon =3r+18$...(i)
When $K_1$ is open and $K_2$ is closed, then current through $12\Omega$ is
$I_2=\frac{\epsilon }{r+12}=\frac{24}{12}=2$ or $\epsilon =2r+24$...(ii)
Solving (i) and (ii), we get $\epsilon =36V,r=6\Omega$
When $K_1$ and $K_2$ are both open, the reading of voltmeter will be the emf of the battery $(=36V)$