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Power dissipated across the $8\,\Omega$ resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the $3\,\Omega$ resistor is :-
(The $1\,\Omega$ and $3\,\Omega$ resistors are in series, and this combination is in parallel with the $8\,\Omega$ resistor.)
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(The $1\,\Omega$ and $3\,\Omega$ resistors are in series, and this combination is in parallel with the $8\,\Omega$ resistor.)
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A
2.0
B
1.0
C
0.5
D
3.0
Detailed Solution
For the $8\,\Omega$ resistor, $P = \dfrac{V^2}{R} \Rightarrow V = \sqrt{PR}$
Voltage drop across $8\,\Omega$ $= \sqrt{2 \times 8} = 4$ V
The branch containing $1\,\Omega$ and $3\,\Omega$ in series is in parallel with the $8\,\Omega$ resistor, so the voltage across this branch is also 4 V.
This 4 V is divided between $1\,\Omega$ and $3\,\Omega$ in the ratio of their resistances (1 : 3).
Therefore the voltage drop across $3\,\Omega$ $= 4 \times \dfrac{3}{1 + 3} = 3$ V
Hence the power dissipated in $3\,\Omega$ $= \dfrac{(3)^2}{3}$
$= 3$ watt
Voltage drop across $8\,\Omega$ $= \sqrt{2 \times 8} = 4$ V
The branch containing $1\,\Omega$ and $3\,\Omega$ in series is in parallel with the $8\,\Omega$ resistor, so the voltage across this branch is also 4 V.
This 4 V is divided between $1\,\Omega$ and $3\,\Omega$ in the ratio of their resistances (1 : 3).
Therefore the voltage drop across $3\,\Omega$ $= 4 \times \dfrac{3}{1 + 3} = 3$ V
Hence the power dissipated in $3\,\Omega$ $= \dfrac{(3)^2}{3}$
$= 3$ watt