Two resistors of 6,Omega and 9,Omega are conn | vedclass

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(A) The resistors are connected in series,so the equivalent resistance is $R_{eq} = 6\,\Omega + 9\,\Omega = 15\,\Omega$.Using Ohm's law,the current $I

Vedclass · Accepted Answer

$384$

Vedclass · Answer

(A) The resistors are connected in series,so the equivalent resistance is $R_{eq} = 6\,\Omega + 9\,\Omega = 15\,\Omega$.Using Ohm's law,the current $I$ flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{120\,V}{15\,\Omega} = 8\,A$.Since the resistors are in series,the same current $I = 8\,A$ flows through the $6\,\Omega$ resistor.The power $P$ consumed by the $6\,\Omega$ resistor is given by the formula $P = I^2 R$.Substituting the values,$P = (8\,A)^2 	imes 6\,\Omega = 64 	imes 6 = 384\,W$.

Two resistors of $6\,\Omega$ and $9\,\Omega$ are connected in series to a $120\,V$ source. The power consumed by the $6\,\Omega$ resistor is ........... $W$.

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