Question

A cell can supply currents of \(1\ \text{A}\) and \(0.5\ \text{A}\) via resistances of \(2.5\ \Omega\) and \(10\ \Omega\) respectively. The internal resistance of the cell is

• A. \(2\ \Omega\)
• B. \(3\ \Omega\)
• C. \(4\ \Omega\)
• D. \(5\ \Omega\)

Hint:

For a cell supplying current through resistance \(R\), always use \(I=\dfrac{E}{R+r}\), where \(r\) is the internal resistance of the cell.

Answer 1

The Correct Option is D

Step 1: Recall the cell equation.
A cell of emf $E$ and internal resistance $r$ drives current $I$ through external resistance $R$ as \[ I = \frac{E}{R + r}, \quad\text{so}\quad E = I(R + r). \]
Step 2: First measurement.
With $R_1 = 2.5\ \Omega$ and $I_1 = 1\ \text{A}$, \[ E = 1(2.5 + r) = 2.5 + r. \]
Step 3: Second measurement.
With $R_2 = 10\ \Omega$ and $I_2 = 0.5\ \text{A}$, \[ E = 0.5(10 + r) = \frac{10 + r}{2}. \]
Step 4: Match the two emf values.
The emf is the same cell, so \[ 2.5 + r = \frac{10 + r}{2}. \]
Step 5: Solve for $r$.
Multiply both sides by $2$: \[ 5 + 2r = 10 + r \quad\Rightarrow\quad r = 5\ \Omega. \]
Step 6: Conclude.
The internal resistance of the cell is $5\ \Omega$ (and back-substituting gives $E = 7.5\ \text{V}$).
\[ \boxed{5\ \Omega} \]