Question

A current is passed in a wire of \( 4 \, \Omega \) resistance by a battery. Find out the internal resistance of the battery, when current is passed in another \( 9 \, \Omega \) resistor by the same battery; while the same heat is produced during the same time in both wires. 
 

Hint:

The internal resistance of a battery affects the total current and power delivered to external resistors.

Answer 1

Let the EMF of the battery be \( E \) and its internal resistance be \( r \). The heat \( H \) is proportional to \( I^2 R \), where \( I \) is the current and \( R \) is the resistance. For both cases, the heat is equal: \[ I_1^2 \cdot 4 = I_2^2 \cdot 9. \] Using Ohm's law: \[ I_1 = \frac{E}{4 + r}, \quad I_2 = \frac{E}{9 + r}. \] Substituting these: \[ \left( \frac{E}{4 + r} \right)^2 \cdot 4 = \left( \frac{E}{9 + r} \right)^2 \cdot 9. \] Simplifying: \[ \frac{4}{(4 + r)^2} = \frac{9}{(9 + r)^2}. \] Taking the square root: \[ \frac{2}{4 + r} = \frac{3}{9 + r}. \] Cross-multiplying: \[ 2(9 + r) = 3(4 + r). \] Expanding and solving: \[ 18 + 2r = 12 + 3r \quad \Rightarrow \quad r = 6 \, \Omega. \]