Question

Two electric bulbs 40 W, 200 V and 100 W, 200 V are connected in series. Then the maximum voltage that can be applied across the combination, without fusing either bulb (in V) is

• A. 280
• B. 400
• C. 300
• D. 200

Answer 1

Answer: (A)

Maximum voltage to a bulb is 200 V. In series, same current passes through each bulb.
Resistance of 40 W bulb = $\frac{200\times 200}{40}=1000\Omega$
Resistance of 100 W bulb = $\frac{200\times 200}{100}=400\Omega$
Current in 40 W bulb for 200 V = $\frac{200}{1000}$ = 0.2 A
Voltage across 100 W bulb for 0.2 A = 400 $\times$ 0.2 = 80 V
Total voltage = 200 + 80 = 280 V