Question

The above battery sends a current \( I_1 \) when \( R = R_1 \) and a current \( I_2 \) when \( R = R_2 \). Obtain the internal resistance of the battery in terms of \( I_1 \), \( I_2 \), \( R_1 \), and \( R_2 \).

Hint:

By applying Ohm’s law to two different scenarios and setting the emf equal in both, we can solve for the internal resistance of the battery using the known values of currents and resistances.

Answer 1

A battery possesses an internal resistance \( r \). When an external resistance \( R_1 \) is connected, the current is \( I_1 \). When the external resistance is changed to \( R_2 \), the current becomes \( I_2 \). The objective is to determine the internal resistance \( r \) as a function of the known values \( I_1 \), \( I_2 \), \( R_1 \), and \( R_2 \).Let the electromotive force (emf) of the battery be denoted by \( \mathcal{E} \). Step 1: Application of Ohm's LawFor the first scenario, with external resistance \( R_1 \) and current \( I_1 \), the total circuit resistance is \( R_1 + r \). Ohm's Law yields:\[\mathcal{E} = I_1 (R_1 + r)\]For the second scenario, with external resistance \( R_2 \) and current \( I_2 \), the total circuit resistance is \( R_2 + r \). Applying Ohm's Law again:\[\mathcal{E} = I_2 (R_2 + r)\] Step 2: Derivation of \( \mathcal{E} \) from both equationsThe emf expressions are:\[\mathcal{E} = I_1 (R_1 + r)\]and\[\mathcal{E} = I_2 (R_2 + r)\] Step 3: Equating the expressions for \( \mathcal{E} \)By setting the two expressions for \( \mathcal{E} \) equal:\[I_1 (R_1 + r) = I_2 (R_2 + r)\] Step 4: Algebraic solution for \( r \)Expanding the equation:\[I_1 R_1 + I_1 r = I_2 R_2 + I_2 r\]Rearranging terms to group \( r \) terms on one side and constant terms on the other:\[I_1 r - I_2 r = I_2 R_2 - I_1 R_1\]Factoring out \( r \):\[r (I_1 - I_2) = I_2 R_2 - I_1 R_1\]Solving for \( r \):\[r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2}\] Final Answer: The internal resistance \( r \) of the battery is calculated as:\[r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2}\]