Power dissipated across the 8 Ω Ω resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the 3 Ω Ω resistor is:-

Text Solution
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Resistances 1 Ω Ω and 3 Ω Ω are connected in series, so effective resistance
R' = 1 + 3 = 4 Ω Ω
Now, R' and 8 Ω Ω are in parallel. We known that potential difference across resistances in parallel order is same

Hence, R' × i 1 = 8i 2
or 4 × i 1 = 8i 2
or i 1 =
i 2 = 2i 2 ......(i)
Power dissipated across 8 Ω Ω resistance is
i 2 2 (8) t = 2W
or i 2 2 t =
= 0.25 W ......(ii)
Power dissipated across 3 Ω Ω resistance is
H = i 1 2 t
= (2i 2 ) 2 t
= 12i 2 2 t
but i 2 2 t = 0.25 W
H = 12 × 0.25 = 3W
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