Two electric bulbs rated 50 W and 100 W are glowing at full power, when used in parallel with a battery of emf 120 V and internal resistance 10Ω. The maximum number of bulbs that can be connected in the circuit when glowing at full power, is:

4

Maximum current possible in bulb = $\frac{50}{100}$ = 0.5 A

Resistance of each bulb $=\frac{V^2}{P}=\frac{100 \times 100}{50}=200 \Omega$

If n be the number of bulb possible, then total resistance of circuit $=\frac{200}{n}$ + 10

Maximum current in the circuit = 0.5 × n

So  $\frac{120}{\frac{200}{n}+10}$ = 0.5 n ⇒ n = 4