137. (A) When two identical bulbs rated '220 V, 60 W' are connected in series to a 220 V mains, the total power consumed is 30 W.
(R) In a series connection, the voltage across each bulb becomes half of the supply voltage.
Key Concept: Electric Power and Energy Consumption
a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that when two identical bulbs rated '220 V, 60 W' are connected in series to a 220 V mains, the total power consumed is 30 W. This is correct because in a series connection, the voltage divides equally, so each bulb gets 110 V. Using $P = \frac{V^2}{R}$, where $R = \frac{(220)^2}{60} = 806.67 \, \Omega$ for each bulb, the power consumed by each bulb at 110 V is $\frac{(110)^2}{806.67} \approx 15 \, \text{W}$. Hence, total power is $15\,\text{W} + 15\,\text{W} = 30\,\text{W}$. The Reason states that in a series connection, the voltage across each bulb becomes half of the supply voltage, which is also true and correctly explains why the power reduces to 30 W.
Your Answer is correct.
a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that when two identical bulbs rated '220 V, 60 W' are connected in series to a 220 V mains, the total power consumed is 30 W. This is correct because in a series connection, the voltage divides equally, so each bulb gets 110 V. Using $P = \frac{V^2}{R}$, where $R = \frac{(220)^2}{60} = 806.67 \, \Omega$ for each bulb, the power consumed by each bulb at 110 V is $\frac{(110)^2}{806.67} \approx 15 \, \text{W}$. Hence, total power is $15\,\text{W} + 15\,\text{W} = 30\,\text{W}$. The Reason states that in a series connection, the voltage across each bulb becomes half of the supply voltage, which is also true and correctly explains why the power reduces to 30 W.