Across the 220 V source of internal resistance $20 \Omega$, how many lamps of $40 \mathrm{~W}, 100 \mathrm{~V}$ can be connected in parallel so that all the lamps may glow with full brightness.

Each lamp is rated $40 \text{ W}$ at $100 \text{ V}$. The resistance of each lamp is $R_L = \dfrac{V^2}{P} = \dfrac{100^2}{40} = \dfrac{10000}{40} = 250 \Omega$.

Let $n$ be the number of lamps connected in parallel. The equivalent resistance of $n$ lamps is $R_p = \dfrac{R_L}{n} = \dfrac{250}{n} \Omega$.

The circuit consists of the source voltage $V_s = 220 \text{ V}$ with internal resistance $r = 20 \Omega$ in series with the parallel combination of lamps. For the lamps to glow with full brightness, the voltage across the parallel combination must be $100 \text{ V}$.

The current flowing through the circuit is $I = \dfrac{V_s - V_L}{r} = \dfrac{220 - 100}{20} = \dfrac{120}{20} = 6 \text{ A}$.

The current through each lamp is $I_L = \dfrac{P}{V_L} = \dfrac{40}{100} = 0.4 \text{ A}$.

The total current is the sum of currents through $n$ lamps, so $I = n \times I_L$.

$6 = n \times 0.4 \Rightarrow n = \dfrac{6}{0.4} = 15$.

Answer: 15