Question

Following figure shows different combinations of identical bulbs connected to identical battery(ies). Which option is correct regarding the total power dissipated in the circuit?

• A. P < Q < R < S
• B. R < Q < P < S
• C. P < Q < R = S
• D. P < R < Q < S

Answer 1

Answer: (D)

. P < R < Q < S.

Let the resistance of each bulb be $R$ and the potential difference of each battery be $V$.

We know that, Power (P) = $\frac{V^2}{R}$

The total resistance in circuit P = 3R

Potential difference in circuit P = V
Therefore, Power in circuit $P=\frac{V^2}{3R}$

The total resistance in circuit Q = R/3

Potential difference in circuit Q = V

Therefore, Power in circuit $Q=\frac{3V^2}{R}$

The total resistance in circuit R = R

Potential difference in circuit R = V
Therefore, Power in circuit $R=\frac{V^2}{R}$

The total resistance in circuit S = R

Potential difference in circuit S = 2V

Therefore Power in circuit $S=\frac{4V^2}{R}$
So the order is S > Q > R > P.