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
$I=\frac{E}{\left(R_1+r\right)}$
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}=63.25\, \Omega$