For a bulb of rating 4V, 6W, we can calculate the resistance using the formula: \[ P = \frac{V^2}{R} \] Thus, \( R = \frac{V^2}{P} = \frac{4^2}{6} = \frac{16}{6} \approx 2.67 \, \Omega \). Now, to operate safely with a 240V mains supply, the total resistance in the series connection must be: \[ R_{{total}} = \frac{V_{{total}}^2}{P_{{total}}} \] For total power 6W per bulb, and for the total power at 240V, the number of bulbs required can be calculated to be 40.
Thus, the correct answer is option (b).
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Two batteries of emf's \(3V \& 6V\) and internal resistances 0.2 Ω \(\&\) 0.4 Ω are connected in parallel. This combination is connected to a 4 Ω resistor. Find:
(i) the equivalent emf of the combination
(ii) the equivalent internal resistance of the combination
(iii) the current drawn from the combination
Name the movement which causes ‘X’ and ‘Y’ to grow downwards and upwards respectively. (Refer above figure)
What is stainless steel? How is it prepared? Write one important property which makes it more useful in making cooking utensils as compared to its primary metal.
What are alloys? How is ‘Brass’ (an alloy) prepared?