For a 4V, 6W bulb, resistance is calculated using: \[ P = \frac{V^2}{R} \] Therefore, \( R = \frac{V^2}{P} = \frac{4^2}{6} = \frac{16}{6} \approx 2.67 \, \Omega \). To operate safely on a 240V supply, the total series resistance must be: \[ R_{{total}} = \frac{V_{{total}}^2}{P_{{total}}} \] With a total power of 6W per bulb and a 240V supply, 40 bulbs are needed.
Thus, the correct answer is option (b).
In the given circuit, the rms value of current (\( I_{\text{rms}} \)) through the resistor \( R \) is: