If power dissipated in the 9 ,Omega resistor | vedclass

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(C) The power dissipated in the $9 \,\Omega$ resistor is given by $P = I_1^2 R_1$. Given $P = 36 \,W$ and $R_1 = 9 \,\Omega$,we have $36 = I_1^2 	ime

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$10$

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(C) The power dissipated in the $9 \,\Omega$ resistor is given by $P = I_1^2 R_1$. Given $P = 36 \,W$ and $R_1 = 9 \,\Omega$,we have $36 = I_1^2 	imes 9$,which gives $I_1^2 = 4$,so $I_1 = 2 \,A$. Since the $9 \,\Omega$ and $6 \,\Omega$ resistors are connected in parallel,the potential difference across them is the same. Let $V_p$ be the potential difference across the parallel combination. Then $V_p = I_1 R_1 = 2 \,A 	imes 9 \,\Omega = 18 \,V$. The current through the $6 \,\Omega$ resistor is $I_2 = \frac{V_p}{R_2} = \frac{18 \,V}{6 \,\Omega} = 3 \,A$. The total current $I$ flowing through the $2 \,\Omega$ resistor is $I = I_1 + I_2 = 2 \,A + 3 \,A = 5 \,A$. The potential difference across the $2 \,\Omega$ resistor is $V_{2\Omega} = I 	imes 2 \,\Omega = 5 \,A 	imes 2 \,\Omega = 10 \,V$.

If power dissipated in the $9 \,\Omega$ resistor in the circuit shown is $36 \,W$,the potential difference across the $2 \,\Omega$ resistor is .......... $V$

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