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 voltagePart (a): Terminal Voltage Across the Cell
Consider a cell with electromotive force (emf) \( E \) and internal resistance \( r \). When connected in series with a variable resistor \( R \), the total resistance in the circuit is \( r + R \). According to Ohm's law, the current \( I \) is: \[I = \frac{E}{r + R}\]The terminal voltage \( V \) across the cell is equivalent to 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 calculated by subtracting the potential drop across the internal resistance from the emf: \[V = E - I r\]Substituting the expression for \( I \): \[V = E - \left( \frac{E}{r + R} \right) r = E \left( \frac{r + R - r}{r + R} \right) = \frac{E R}{r + R}\]This result validates the derived expression for terminal voltage.% Analyze the behavior of terminal voltageAnalyzing \( V \) as \( R \) changes 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, resulting in the entire emf being dissipated 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, making the voltage drop across the internal resistance (\( I r \)) negligible.% Describe the graphical variation for terminal voltageThe graph of \( V \) versus \( R \) begins at \( V = 0 \) for \( R = 0 \). It rises steeply initially, then more gradually, and asymptotically approaches \( V = E \) as \( R \) becomes very large. The shape of the curve is hyperbolic growth, consistent with the equation \( V = \frac{E R}{r + R} \).% Derivation of currentPart (b): Current Supplied by the Cell
The current supplied by the cell is identical to the current flowing through the circuit: \[I = \frac{E}{r + R}\]% Analyze the behavior of currentAnalyzing the behavior of \( I \) as \( R \) varies:- When \( R = 0 \): \[I = \frac{E}{r + 0} = \frac{E}{r}\]This represents the maximum current, occurring during 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 diminishes to zero as the total resistance of the circuit increases significantly.% Describe the graphical variation for currentThe graph of \( I \) versus \( R \) starts at \( I = \frac{E}{r} \) for \( R = 0 \). It decreases rapidly at first, then more slowly, and asymptotically approaches \( I = 0 \) as \( R \) becomes very large. The curve exhibits a hyperbolic decay shape, reflecting the relationship \( I = \frac{E}{r + R} \).