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

Internal resistance is not changed with external resistor. ( So we first calculate internal resistance .), \[\begin{align*} r&=\left(\frac{\mathcal{E}-V}{V}\right)\times R\\ &=\frac{1.5-1.25}{1.25}\times 25\\ \therefore r&=5 \hspace{0.1cm} \Omega\\ \end{align*}\] Now, \[\begin{align*} I&=\frac{\mathcal{E}}{R+r}\\ &=\frac{1.5}{10+5}\\ \therefore I&=0.1 \hspace{0.1cm} A\\ \end{align*}\] Again, \[\begin{align*} V&=IR\\ &=0.1 \times 10\\ \therefore V&=1 \hspace{0.1cm} V\\ \end{align*}\] 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.