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The total power dissipated in watt in the circuit shown here is

Asked in CBSE AIPMT 2007 · Heating and power in resistors

Figure: Heating and power in resistors

(1) 40
(2) 54
(3) 4
(4) 16

Answer: (2) 54

Step-by-step solution

6 Ω∥3 Ω=(18)/9=2 Ω.

In series with 4 Ω: R=6 Ω.

P=(V²)/R=(18²)/6=(324)/6=54 W.

Why the other options are wrong

(1) No single slip gives this value; 40 W would need a total resistance of 8.1 Ω, and the network gives 6 Ω.
(3) No single slip gives this value; 4 W would need 81 Ω in total, far more than the three resistors can make.
(4) No single slip gives this value; 16 W would need about 20 Ω in total, more than even all three in series (13 Ω).

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