Question

The emf of a cell is $\varepsilon$ 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. $\varepsilon=1\, V,\, r=1\, \Omega$
• B. $\varepsilon=2\, V,\, r=1\, \Omega$
• C. $\varepsilon=2\, V,\, r=2\, \Omega$
• D. $\varepsilon=2.5\, V,\, r=0.5\, \Omega$

Answer 1

Answer: (B)

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