Question

The emf of a cell is $\epsilon$ and its internal resistance is $r.$ Its terminals are connected to a resistance $R$. The potential difference between the terminals is $1.6V$ for $R=4\Omega$, and $1.8V$ for $R=9\Omega$. Then,

• A. $\epsilon =1V,r=1\Omega$
• B. $\epsilon =2V,r=1\Omega$
• C. $\epsilon =2V,r=2\Omega$
• D. $\epsilon =2.5V,r=0.5\Omega$

Answer 1

Answer: (B)

Current in the circuit, $I=\frac{\epsilon }{R+r}$
p.d. across cell $=$ p.d. across $R=IR=\frac{\epsilon R}{R+r}$
Set up two equations with the given data $1.6=\frac{4\epsilon }{4+r}$...(i)
$1.8=\frac{9\epsilon }{9+r}$...(ii)
Solving equations (i) and (ii), we get $r=1\Omega ,\epsilon =2V$