Question

A set up is such that there are three similar resistors, each of 20 ohms resistance. Two of them are connected in parallel, and this combination is connected in series with the third one. The maximum power that can be consumed by each resistor is 30 W. Then, what is the maximum power that can be consumed by the combination of all three resistors?

(a) 30

(b) 20

(c) 35

(d) 45

This question was addressed to me in an online interview.

This intriguing question originated from Combination of Resistors topic in division Current Electricity of Physics – Class 12

Answer 1

Right answer is (d) 45

The explanation is: The equivalent overall resistance of the parallel combination is:

 \(\frac {1}{R1} = \frac {1}{20} + \frac {1}{20} = \frac {2}{20} = \frac {1}{10}\) → R1 = 10 ohms.

 R1 is in series with R2; So, R3 = R1 + R2 = 10 + 20 = 30 ohms.

Now, we can employ the method of cross-multiplication:

For 20 ohms resistor  → 30 W power consumed

For 30 ohms resistor combination → x

20x = 30 × 30

x = \(\frac {30 \times 30}{20}\)

x = 45

Therefore, the power consumed by the parallel combination is 45 ohms.