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Two resistors A (3 ?) and B (6 ?) are connected in parallel. This combination is connected in series to two other resistors C (3 ?) and D (6 ?) and a 5.5 V battery of negligible internal resistance. Let PB and PD be the rate of energy dissipated in resistors B and D, respectively. Then the ratio (PB / PD) is:

A.
\(\frac{1}{3}\)
B.
\(\frac{1}{9}\)
C.
\(\frac{4}{9}\)
D.
\(\frac{2}{3}\)

Solution:

The correct answer is 1/9.Concept:Resistance in series:When resistors are connected end-to-end in a circuit, they are said to be in "series."R = R1 + R2 + R3 +....+ RnCurrent remains the same across all the resistors connected in series. Resistance in parallel: When both ends of resistors are connected, they are said to be in parallel. 1/R = 1/R1 + 1/R2 = 1/R3 ....The potential difference across all the resistors connected in parallel remains the same. Electric Power:The energy consumed per unit of time or the rate of doing work is called power. Power = work done / time taken = Vq / t = VI = V2/R = I2RExplanation:Resistors A (3 ?), and B (6 ?) are connected in parallel.Equivalent resistance, 1/R1 = 1/3 + 1/6 = 2+1/ 6 = 3/6 = 1/2 ? R1 = 2 ? Now, the resistors R1, C, and D are in series.R2 = 2 + 3 + 6 = 11 ? Equivalent current, I in the circuit = V/R2 = 5.5 / 11 = 0.5 = 1/ 2 A.Potential difference across B = IR1 = 1/2 À” 2 = 1 VPB = V2/ D = 1 À” 1 / 6 = 1/6 Watt (since current remains same in series combination)PD = I2B = 1/2 À” 1/2 À” 6 = 3/2 Watt (since voltage remains the same in parallel combination)The ratio (PB/PD) = 1/6 À” 2/3 = 1/ 9.

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