Question

In a domestic circuit five LED bulbs are arranged as shown. The source voltage is 220 V and the power rating of each bulb is marked in the circuit diagram. Based on the following circuit diagram, answer the following questions:

(a) State what happens when
(i) key K₁ is closed.
(ii) key K₂ is closed.
(b) Find the current drawn by the bulb B when it glows.
(c) Calculate
(i) the resistance of bulb B, and
(ii) total resistance of the combination of four bulbs B, C, D and E.
OR
(c) What would happen to the glow of all the bulbs in the circuit when keys K₁ and K₂ both are closed and the bulb C suddenly get fused ? Give reason to justify your answer.

Answer 1

(i) When \(K_1\) is closed:
The 22 W, 220 V bulb forms a complete circuit and operates at its rated voltage.
    The current through this bulb is calculated as \(I = \frac{P}{V} = \frac{22 \, \text{W}}{220 \, \text{V}} = 0.1 \, \text{A}\).
    This bulb glows normally. The other four bulbs remain \textbf{OFF} because \(K_2\) is open, breaking their circuit.
(ii) When \(K_2\) is closed:
   The four 11 W, 55 V bulbs are connected in series across the 220 V source.
   In this series configuration, the voltage divides equally across the four identical bulbs.
   The voltage across each bulb is: \[V_{each bulb} = \frac{220V}{4} = 55 \, V\]
   Since each bulb receives its rated voltage of 55 V, they all glow normally.
   The current through each bulb (and the series circuit) is calculated using its power rating:
   \[I = \frac{P}{V} = \frac{11}{55} = 0.2 \, A\]
   Therefore, a current of 0.2 A flows through the entire series circuit.
Summary of observations:
   When \(K_1\) is closed: The 22 W bulb glows normally with 0.1 A. The 11 W bulbs are off.
   When \(K_2\) is closed: The four 11 W bulbs glow normally, each at 55 V, with a total series current of 0.2 A.