Question

When a cell is connected across a resistance $R$, the current through it is 0.7 A and when the same cell is connected across a resistance $2R$, the current through it is 0.42 A. If the same cell is connected across a resistance $3R$, the current through it is

• A. 0.4 A
• B. 0.3 A
• C. 0.25 A
• D. 0.35 A

Hint:

- Use the formula $I = E/(R + r)$ for series combination of internal resistance.
- Solve two simultaneous equations to find the internal resistance $r$.
- Once $r$ is known, easily calculate current for any other external resistance.
- Always check units consistency: ohms, amperes.
- Plotting a simple $I$ vs $R$ graph can help visualize the effect of internal resistance.

Answer 1

The Correct Option is B

1. Let the emf of the cell be $E$ and internal resistance $r$. By Ohm's law, $I = \frac{E}{R + r}$.
2. For $R$, $I_1 = 0.7 = \frac{E}{R + r} \Rightarrow E = 0.7(R + r)$
3. For $2R$, $I_2 = 0.42 = \frac{E}{2R + r} \Rightarrow E = 0.42(2R + r)$
4. Equate $E$: $0.7(R + r) = 0.42(2R + r)$
\[ 0.7R + 0.7r = 0.84R + 0.42r \Rightarrow 0.28r = 0.14R \Rightarrow r = 0.5 R \]
5. For $3R$, $I_3 = \frac{E}{3R + r} = \frac{0.7(R + r)}{3R + r} = \frac{0.7(1.5 R)}{3.5 R} = 0.3~\text{A}$