Question

A battery of emf E and internal resistance r is connected to an external resistance R.
(I) Expression for current I and maximum current:.
(II) Terminal voltage V and its maximum value:.

Answer 1

Expression for Current, Terminal Voltage, and Internal Resistance

(I) Expression for Current \( I \) and Maximum Current

By Ohm's Law, the total resistance in the circuit is \( R + r \), where \( R \) is the external resistance and \( r \) is the internal resistance.

The current in the circuit is given by:

\[ I = \frac{E}{R + r} \]

where \( E \) is the electromotive force (EMF) of the battery.

The maximum current occurs when the external resistance \( R = 0 \), and is given by:

\[ I_{\text{max}} = \frac{E}{r} \]

(II) Terminal Voltage \( V \) and its Maximum Value

The terminal voltage \( V \) is the potential difference across the external resistance, and is given by:

\[ V = E - I r = E - \frac{E r}{R + r} = \frac{E R}{R + r} \]

The maximum terminal voltage occurs when the current is minimum (i.e., when \( R \to \infty \), meaning no current flows). In this case, the terminal voltage becomes equal to the EMF of the battery:

 

\[ V_{\text{max}} = \lim_{R \to \infty} \frac{E R}{R + r} = E \]

(III) Expression for Internal Resistance in Terms of \( I_1, I_2, R_1, R_2 \)

Given the following expressions for currents \( I_1 \) and \( I_2 \) with resistances \( R_1 \) and \( R_2 \), respectively:

\[ I_1 = \frac{E}{R_1 + r}, \quad I_2 = \frac{E}{R_2 + r} \]

Cross-multiply the above equations: \[ I_1 (R_1 + r) = E, \quad I_2 (R_2 + r) = E \] Equating the two expressions: \[ I_1 (R_1 + r) = I_2 (R_2 + r) \] Expanding and simplifying: \[ I_1 R_1 + I_1 r = I_2 R_2 + I_2 r \] \[ I_1 r - I_2 r = I_2 R_2 - I_1 R_1 \] Factor 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 expression for the internal resistance \( r \) is: \[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]