Question:medium

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:.

Updated On: Jan 13, 2026
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Solution and Explanation

Current, Terminal Voltage, and Internal Resistance Formulas

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

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

The circuit current is calculated as:

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

Here, \( E \) represents the electromotive force (EMF) of the battery.

Maximum current is achieved when external resistance \( R = 0 \), yielding:

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

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

Terminal voltage \( V \) is the potential difference across the external resistance, defined as:

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

The maximum terminal voltage occurs at minimum current (i.e., when \( R \to \infty \), signifying no current flow). In this scenario, terminal voltage equals the battery's EMF:

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

(III) Internal Resistance Formula using \( I_1, I_2, R_1, R_2 \)

Given current expressions \( I_1 \) and \( I_2 \) for resistances \( R_1 \) and \( R_2 \), respectively:

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

Cross-multiplication yields: \[ I_1 (R_1 + r) = E, \quad I_2 (R_2 + r) = E \] Equating these gives: \[ I_1 (R_1 + r) = I_2 (R_2 + r) \] Expanding and simplifying: \[ I_1 R_1 + I_1 r = I_2 R_2 + I_2 r \] Rearranging terms: \[ I_1 r - I_2 r = I_2 R_2 - I_1 R_1 \] 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 \( r \) is calculated as: \[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]
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