Question

A cell of emf $E$ and internal resistance $r$ supplies currents for the same time $t$ through external resistance $R_{1}= 100 \,\Omega $ and $ R_{2}=40 \, \Omega $ separately. If the heat developed in both cases is the same, then the internal resistance of the cell is given by

• A. $28.6\, \Omega $
• B. $70 \, \Omega $
• C. $63.3 \, \Omega$
• D. $140 \, \Omega $

Answer 1

Answer: (C)

Current drawn from the cell in resistance $R_{1}$ will be
$I=E \left(\right.R_{1}+r\left.\right)$
Therefore, the heat produced in $R_{1}$ i.e.
$H_{1}=\frac{E^{2} R_{1} t}{\left(R_{1} + r\right)^{2}}$
Heat produced in $R_{2} , H_{2}=\frac{E^{2} R_{2} t}{\left(R_{2} + r\right)^{2}}$
As per question $H_{1}=H_{2}$
or $\frac{E^{2} R_{1} t}{\left(R_{1} + r\right)^{2}}=$ $\frac{E^{2} R_{2} t}{\left(R_{2} + r\right)^{2}}$
On solving we get;
$r=\sqrt{R_{1} R_{2}}$
$=\sqrt{100 \times 40}=\text{63.25 }\,\Omega $