Question

A cell of emf \( E \) and internal resistance \( r \) is connected across a resistor of variable resistance \( R \). Show graphically the variation of
(a) the terminal voltage across the cell,
(b) the current supplied by the cell,
with \( R \) as it is increased from 0 to the maximum value.

Hint:

When analyzing a circuit with internal resistance, note that the terminal voltage \( V \) approaches the emf \( E \) as the external resistance increases, while the current \( I \) decreases inversely with the total resistance. The shapes of the graphs are determined by the denominator \( r + R \).

Answer 1

Derivation of terminal voltage Part (a): Terminal Voltage Across the Cell
Consider a cell with emf \( E \) and internal resistance \( r \), connected in series with a variable resistor \( R \). The total resistance in the circuit is \( r + R \). The current \( I \) in the circuit is given by Ohm’s law applied to the entire circuit: \[ I = \frac{E}{r + R} \] The terminal voltage \( V \) across the cell is the voltage across the resistor \( R \): \[ V = I R = \left( \frac{E}{r + R} \right) R = \frac{E R}{r + R} \] Alternatively, the terminal voltage can be found by considering the potential drop across the internal resistance: \[ V = E - I r \] Substitute \( I \): \[ V = E - \left( \frac{E}{r + R} \right) r = E \left( \frac{r + R - r}{r + R} \right) = \frac{E R}{r + R} \] This confirms our expression for the terminal voltage. % Analyze the behavior of terminal voltage Now, analyze the behavior of \( V \) as \( R \) varies from 0 to a very large value:
- When \( R = 0 \): \[ V = \frac{E \cdot 0}{r + 0} = 0 \] The terminal voltage is zero because the cell is short-circuited, and the entire emf is dropped across the internal resistance.
- When \( R = r \): \[ V = \frac{E r}{r + r} = \frac{E r}{2r} = \frac{E}{2} \] - As \( R \to \infty \): \[ V \to \frac{E R}{R} = E \] The terminal voltage approaches the emf \( E \), as the current becomes very small, and the voltage drop across the internal resistance (\( I r \)) becomes negligible. % Describe the graphical variation for terminal voltage The graph of \( V \) versus \( R \) starts at \( V = 0 \) when \( R = 0 \), increases rapidly at first, then more gradually, and asymptotically approaches \( V = E \) as \( R \) becomes very large. The curve is a hyperbolic growth shape, reflecting the form of the equation \( V = \frac{E R}{r + R} \). % Derivation of current Part (b): Current Supplied by the Cell
The current supplied by the cell is the same as the current in the circuit: \[ I = \frac{E}{r + R} \] % Analyze the behavior of current Analyze the behavior of \( I \) as \( R \) varies:
- When \( R = 0 \): \[ I = \frac{E}{r + 0} = \frac{E}{r} \] This is the maximum current, corresponding to a short circuit.
- When \( R = r \): \[ I = \frac{E}{r + r} = \frac{E}{2r} \] - As \( R \to \infty \): \[ I \to \frac{E}{R} \to 0 \] The current approaches zero as the total resistance becomes very large. % Describe the graphical variation for current The graph of \( I \) versus \( R \) starts at \( I = \frac{E}{r} \) when \( R = 0 \), decreases rapidly at first, then more slowly, and asymptotically approaches \( I = 0 \) as \( R \) becomes very large. The curve is a hyperbolic decay shape, reflecting the form of the equation \( I = \frac{E}{r + R} \).