JEE Main 2017 — Current Electricity Question with Solution

Assume equivalent resistance of the infinite network is x.

The circuit can be redrawn like this,


Here 4 $$\Omega$$ and x $$\Omega$$ are parallel,

$$\therefore$$ $$\frac{1}{x_{eq}}$$ = $$\frac{1}{4}$$ + $$\frac{1}{x}$$

$$\Rightarrow$$ x_eq = $$\frac{4x}{4+x}$$

$$\therefore$$ Equivalent resistance between point A and B is ,

x = $$\frac{4x}{4+x}$$ + 2

$$\Rightarrow$$ x = $$\frac{8+6x}{4+x}$$

$$\Rightarrow$$ 4x + x² = 8 + 6x

$$\Rightarrow$$ x² $$-$$ 2x $$-$$ 8 = 0

$$\Rightarrow$$ x = $$\frac{2\pm \sqrt{4-4\left(1\right)\left(-8\right)}}{2}$$

$$\Rightarrow$$x = $$\frac{2\pm 6}{2}$$

$$\Rightarrow$$ = 4 $$\Omega$$

(as negative value of x is not acceptable)

$$\therefore$$ Reading of Aonmeter, A₁ = $$\frac{V}{R+r}$$

A₁ = $$\frac{9}{4+0.5}$$ = 2 A