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The total power dissipated in watts in the circuit shown here is
($6\,\Omega$ and $3\,\Omega$ are in parallel, and this combination is in series with $4\,\Omega$ across an 18 V battery.)
($6\,\Omega$ and $3\,\Omega$ are in parallel, and this combination is in series with $4\,\Omega$ across an 18 V battery.)
A
40
B
54
C
4
D
16
Detailed Solution
The $6\,\Omega$ and $3\,\Omega$ resistors are in parallel: $R_p = \dfrac{6 \times 3}{6 + 3} = \dfrac{18}{9} = 2\,\Omega$
This is in series with the $4\,\Omega$ resistor: $R = 2 + 4 = 6\,\Omega$
Power dissipated: $P = \dfrac{V^2}{R}$
$P = \dfrac{(18)^2}{6} = \dfrac{324}{6}$
$P = 54$ W
This is in series with the $4\,\Omega$ resistor: $R = 2 + 4 = 6\,\Omega$
Power dissipated: $P = \dfrac{V^2}{R}$
$P = \dfrac{(18)^2}{6} = \dfrac{324}{6}$
$P = 54$ W