In the given circuit,the current in 8 Omega r | vedclass

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(D) Let the current in the $8 \Omega$ resistor be $I_1 = 1.5 	ext{ A}$.The potential difference $(V_p)$ across the $8 \Omega$ resistor is $V_p = I_1

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$5.5$

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(D) Let the current in the $8 \Omega$ resistor be $I_1 = 1.5 	ext{ A}$.The potential difference $(V_p)$ across the $8 \Omega$ resistor is $V_p = I_1 	imes R_1 = 1.5 	ext{ A} 	imes 8 \Omega = 12 	ext{ V}$.Since the $3 \Omega$ resistor is in parallel with the $8 \Omega$ resistor,the potential difference across it is also $12 	ext{ V}$.The current $(I_2)$ flowing through the $3 \Omega$ resistor is $I_2 = \frac{V_p}{R_2} = \frac{12 	ext{ V}}{3 \Omega} = 4 	ext{ A}$.The total current $(I)$ flowing in the circuit is the sum of the currents in the parallel branches: $I = I_1 + I_2 = 1.5 	ext{ A} + 4 	ext{ A} = 5.5 	ext{ A}$.

In the given circuit,the current in $8 \Omega$ resistance is $1.5 \text{ A}$. The total current $(I)$ flowing in the circuit is (in $\text{ A}$)

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