Current Electricity • NEET Revision

Example 3.35

EMF & Internal Resistance

Question

A battery of emf 2 V and internal resistance r is connected in series with a resistor of 10 Ω through an ammeter of resistance 2 Ω. The ammeter reads 50 mA. Draw the circuit diagram and calculate the value of r.

Given

EMF \(E = 2\ \text{V}\)
External resistor \(R = 10\ \Omega\)
Ammeter resistance \(R_A = 2\ \Omega\)
Current \(I = 50\ \text{mA} = 0.05\ \text{A}\)

Asked

Internal resistance \(r\) (with circuit diagram — shown above).

Formula

Series circuit: \(E = I\,(R + R_A + r)\)

Solution

All elements are in series, so total resistance \(= R + R_A + r = 12 + r\).
Apply \(E = I(R + R_A + r)\): \(\;2 = 0.05\,(12 + r)\).
\(12 + r = \dfrac{2}{0.05} = 40\).
\(r = 40 - 12 = 28\ \Omega\).

\(r = 28\ \Omega\)

Short Cut

Total circuit resistance \(= \dfrac{E}{I} = \dfrac{2}{0.05} = 40\ \Omega\). Subtract everything you can see: \(r = 40 - 10 - 2 = 28\ \Omega\). Never forget the ammeter's own resistance!