Solveeit Logo
Login

Question

Question: Two electric bulbs whose resistances are in the ratio of 1:2 are connected in series. The powers dis...

Two electric bulbs whose resistances are in the ratio of 1:2 are connected in series. The powers dissipated in them have the ratio

A

1 : 2

B

2 : 1

C

1 : 1

D

1 : 4

Answer

1 : 2

Explanation

Solution

In series, current is same in both the bulbs, hence P∝R(P=i2R)P \propto R(P = i^{2}R) ∴P1P2=R1R2=12\therefore\frac{P_{1}}{P_{2}} = \frac{R_{1}}{R_{2}} = \frac{1}{2}