Question

In the circuit shown in figure. (1), R₃ is a variable resistance. As the value R₃ is changed, current I though the cell varies as shown. Obviously, the variation is asymptotic, i.e. I → 6A as R₃→ ∞. Resistances R₁ and R₂ are, respectively –

• A. 4 Ω, 2 Ω
• B. 2 Ω, 4 Ω
• C. 2 Ω, 2 Ω
• D. 1 Ω, 4 Ω

Answer 1

Answer: (C)

2 Ω, 2 Ω

Answer 2

For R₃ = 0, I = 9A (from the graph).In this situation, the circuit can be drawn as shown in figure (1).

Here I = $\frac{36}{R_{1} + R_{2}}$ = 9

or R₁ + R₂ = 4 …(1)

For R₃ → ∞ , equivalent resistance of R₂ and R₃ in parallel will be

$\frac{1}{R_{eq}} = \frac{1}{R_{2}} + \frac{1}{R_{3} \rightarrow \infty}$

= $\frac{1}{R_{2}}$

[Current through R₂ will be zero because we have a short circuit (R₃ = 0) across R₂]

or R_eq = R₂

In this situation, the circuit can be drawn as shown figure (2).

Now I = $\frac{36}{R_{1} + 2R_{2}}$

= 6 (given)

∴ R₁ + 2R₂ = 6 …(2)

from eqs. (1) and (2)

R₁ = 2Ω, R₂ = 2Ω