A battery of emf 1.5 V has a terminal potential difference of 1.25 V when a resistor of 25 Ω is joined to it. Calculate the current flowing, the internal resistance and terminal potential difference when a resistance of 10 Ω replaces 25 Ω resistor.

A battery of emf 1.5 V has a terminal potential difference of 1.25 V when a resistor of 25 $\mathrm{\Omega }$ is joined to it. Calculate the current flowing, the internal resistance and terminal potential difference when a resistance of 10 $\mathrm{\Omega }$ replaces 25 $\mathrm{\Omega }$ resistor.
Solution:
Given,
emf , $\mathcal{E}$ = 1.5 V
terminal potential difference, $V$ = 1.25 V
resistor, $R$ = 25 $\mathrm{\Omega }$
current, $I$ = ?
at $R$ = 10 $\mathrm{\Omega }$ internal resistance, $r$ = ?
terminal potential difference, $V$ = ? at $R$ = 10 $\mathrm{\Omega }$

Internal resistance is not changed with external resistor. ( So we first calculate internal resistance .), $$\begin{array}{rl}r& =\left(\frac{\mathcal{E}-V}{V}\right)\times R\\ & =\frac{1.5-1.25}{1.25}\times 25\\ \therefore r& =5\mathrm{\Omega }\end{array}$$ Now, $$\begin{array}{rl}I& =\frac{\mathcal{E}}{R+r}\\ & =\frac{1.5}{10+5}\\ \therefore I& =0.1A\end{array}$$ Again, $$\begin{array}{rl}V& =IR\\ & =0.1\times 10\\ \therefore V& =1V\end{array}$$ Hence, the current flowing is 0.1 A, internal resistance is 5 Ω and terminal potential difference is 1 V when 10 Ω resistor replaces 25 Ω resistor.