The current in a simple series circuit is 5.0 | vedclass

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(B) Let the original resistance of the circuit be $R$ and the voltage of the source be $V$.According to Ohm's law, $V = I 	imes R$.In the first case,

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$8$

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(B) Let the original resistance of the circuit be $R$ and the voltage of the source be $V$.According to Ohm's law, $V = I 	imes R$.In the first case, $V = 5.0 	imes R$.In the second case, when an additional resistance of $2.0 \, \Omega$ is added, the total resistance becomes $(R + 2.0) \, \Omega$ and the current becomes $4.0 \, A$.So, $V = 4.0 	imes (R + 2.0)$.Since the voltage $V$ remains constant, we equate the two expressions:$5.0 	imes R = 4.0 	imes (R + 2.0)$$5R = 4R + 8.0$$5R - 4R = 8.0$$R = 8.0 \, \Omega$.Therefore, the original resistance of the circuit was $8 \, \Omega$.

The current in a simple series circuit is $5.0 \, A$. When an additional resistance of $2.0 \, \Omega$ is inserted, the current drops to $4.0 \, A$. The original resistance of the circuit in $\Omega$ was:

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