The potential drop across the 3 ,Omega resist | vedclass

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(A) First, calculate the equivalent resistance of the circuit. The $3 \,\Omega$ and $6 \,\Omega$ resistors are in parallel, so their equivalent resist

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$1$

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(A) First, calculate the equivalent resistance of the circuit. The $3 \,\Omega$ and $6 \,\Omega$ resistors are in parallel, so their equivalent resistance $R_p$ is given by:$R_p = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2 \,\Omega$This parallel combination is in series with the $4 \,\Omega$ resistor. Therefore, the total equivalent resistance $R_{eq}$ of the circuit is:$R_{eq} = 4 \,\Omega + 2 \,\Omega = 6 \,\Omega$The total current $i$ flowing from the $3 \,V$ battery is:$i = \frac{V_{total}}{R_{eq}} = \frac{3 \,V}{6 \,\Omega} = 0.5 \,A$The potential drop across the parallel combination (which includes the $3 \,\Omega$ resistor) is:$V_{parallel} = i 	imes R_p = 0.5 \,A 	imes 2 \,\Omega = 1 \,V$Since the $3 \,\Omega$ and $6 \,\Omega$ resistors are connected in parallel, the potential drop across each of them is the same, which is $1 \,V$.

The potential drop across the $3 \,\Omega$ resistor is ............... $V$.

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