Three resistors of 4,Omega,6,Omega,and 12,Ome | vedclass

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(C) The equivalent resistance $R_p$ of the resistors $4\,\Omega$,$6\,\Omega$,and $12\,\Omega$ connected in parallel is given by:$\frac{1}{R_p} = \frac

Vedclass · Accepted Answer

$0.25$

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(C) The equivalent resistance $R_p$ of the resistors $4\,\Omega$,$6\,\Omega$,and $12\,\Omega$ connected in parallel is given by:$\frac{1}{R_p} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = \frac{3+2+1}{12} = \frac{6}{12} = \frac{1}{2} \Rightarrow R_p = 2\,\Omega$.The total resistance of the circuit including the internal resistance $r = 1\,\Omega$ is $R_{total} = R_p + r = 2\,\Omega + 1\,\Omega = 3\,\Omega$.The total current $I$ flowing from the battery is $I = \frac{V}{R_{total}} = \frac{1.5\,V}{3\,\Omega} = 0.5\,A$.Since the resistors are in parallel,the voltage across each resistor is the same,equal to the voltage across the parallel combination $V_p = I 	imes R_p = 0.5\,A 	imes 2\,\Omega = 1.0\,V$.The current $I_1$ through the $4\,\Omega$ resistor is $I_1 = \frac{V_p}{4\,\Omega} = \frac{1.0\,V}{4\,\Omega} = 0.25\,A$.The rate of Joule heating (power) in the $4\,\Omega$ resistor is $P = I_1^2 	imes R = (0.25\,A)^2 	imes 4\,\Omega = 0.0625 	imes 4 = 0.25\,W$.

Three resistors of $4\,\Omega$,$6\,\Omega$,and $12\,\Omega$ are connected in parallel,and the combination is connected in series with a $1.5\,V$ battery of $1\,\Omega$ internal resistance. The rate of Joule heating in the $4\,\Omega$ resistor is ................ $W$.

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