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 1.8 V 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

The correct option is B $\epsilon =2V,r=1\Omega$

The current in the circuit is $I=\frac{\epsilon }{R+r}$
The terminal voltage , $V=IR=\frac{\epsilon R}{R+r}$
using given data, $1.6=\frac{4\epsilon }{4+r}\Rightarrow 8+2r=5\epsilon ...\left(1\right)$
and $1.8=\frac{9\epsilon }{9+r}\Rightarrow 9+r=5\epsilon ...\left(2\right)$
$\left(1\right)-\left(2\right)\Rightarrow (8-9)-(2r-r)=0$ or $-1+r=0$ or $r=1\Omega$
using (1), $5\epsilon =8+2\left(1\right)=10\Rightarrow \epsilon =2V$