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A current of 3 amp. flows through the $2\,\Omega$ resistor shown in the circuit. The power dissipated in the $5\,\Omega$ resistor is -
(Three branches are connected in parallel across the battery: $2\,\Omega$; $4\,\Omega$; and $1\,\Omega$ in series with $5\,\Omega$.)
(Three branches are connected in parallel across the battery: $2\,\Omega$; $4\,\Omega$; and $1\,\Omega$ in series with $5\,\Omega$.)
A
1 watt
B
5 watt
C
4 watt
D
2 watt
Detailed Solution
The three branches ($2\,\Omega$; $4\,\Omega$; and $1\,\Omega + 5\,\Omega$) are in parallel, so the voltage across all three branches is the same.
Voltage across the $2\,\Omega$ resistor: $V = 3 \times 2 = 6$ V
Let $i$ be the current in the branch containing $1\,\Omega$ and $5\,\Omega$ in series.
$3 \times 2 = i \times (5 + 1)$
$i = \dfrac{6}{6} = 1$ A
Power dissipated in the $5\,\Omega$ resistor: $P = i^2 R$
$P = (1)^2 \times 5 = 5$ watt
Voltage across the $2\,\Omega$ resistor: $V = 3 \times 2 = 6$ V
Let $i$ be the current in the branch containing $1\,\Omega$ and $5\,\Omega$ in series.
$3 \times 2 = i \times (5 + 1)$
$i = \dfrac{6}{6} = 1$ A
Power dissipated in the $5\,\Omega$ resistor: $P = i^2 R$
$P = (1)^2 \times 5 = 5$ watt