First slide

Electrical power

Question

Three equal resistances each of 10 $Ω$ are connected as shown in the fig. The maximum power consumed
by each resistor is 20 W. Then maximum power consumed by the combination is

Easy

A

60W
B

30w
C

15W
D

13.3W

Solution

We know that
$P_{p} = P_{1} + P_{2} and P_{s} = (\frac{P_{1} P_{2}}{P_{1} + P_{2}})$
Resistances R₁ and R₂, are connected in parallel. Hence their equivalent power is (20 + 20) = 40 W. Now R₃ is it in series and hence
$P_{s} = \frac{20 \times 40}{20 + 40} = \frac{800}{60} = 13 \cdot 3 W$

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