Question

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

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

Answer 1

Answer: (B)

$ϵ=2V,r=1\Omega$

The current flowing through $R_1=4\Omega$ is $I_1=\frac{\epsilon }{R_1+r}$ and potential difference across cell $=$ potential difference across $R_1$.
Hence,
$I_1{R}_1=\left(\frac{\epsilon }{R_1+r}\right)R_1$
$\Rightarrow 1.6=\left(\frac{\epsilon }{4+r}\right)\times 4$
$\therefore \epsilon =1.6+0.4r$ ................................(1)
Similarly, the current flowing through $R_2=9\Omega$ is $I_2=\frac{\epsilon }{R_2+r}$ and potential difference across cell $=$ potential difference across $R_2$.
Hence,
$I_2{R}_2=\left(\frac{\epsilon }{R_2+r}\right)R_2$
$\Rightarrow 1.8=\left(\frac{\epsilon }{9+r}\right)\times 9$
$\epsilon =(9+r)\times 0.2$
From (1)
$1.6+0.4r=(9+r)\times 0.2$
$1.6+0.4r=1.8+0.2r$
$0.2r=0.2$
$\therefore r=1\Omega$
Using this value in (1) we get
$\epsilon =1.6+0.4\left(1\right)$
$\therefore \epsilon =2$ V