Question

The value of the resistance $R$ in the circuit shown below so that electric bulb consumes the rated power is

• A. $4\Omega$
• B. $6\Omega$
• C. $18\Omega$
• D. $8\Omega$

Answer 1

The correct option is A $4\Omega$

If power = V I

0.5 = 3$\times$ I

I=$\frac{0.5}{3}A$

If we want bulb to consume rated power , then current through bulb should be $\frac{0.5}{3}A$

$\to$ bulb and resistance are in parallel so voltage across resistance should be 3 v

3 = $I_{1}R$ (i)

Power = $\frac{V^2}{R}$

0.5 = $\frac{9}{R}$ ,R = 18$\Omega$ Bulb's resistance

$\to$ bulb resistance are in parallel

$IR=I_1{R}_1\frac{05}{3}\times 18=I_1{I}_1$ (i)

net current $=\frac{6V}{0.5+4+}I+I_1=\frac{6V}{0.5+4+\frac{18R}{18+R}}\frac{05}{3}+\frac{3}{R}=\frac{6(18+R)}{9+\frac{R}{2}+72+4R}$

On solving equation , we get R = 4$\Omega$