In the circuit shown,the potential difference | vedclass

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(C) The circuit consists of a voltage source of $48 \, V$ connected in series with resistors of $100 \, \Omega$,$80 \, \Omega$,and $20 \, \Omega$. Fir

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$4.8$

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(C) The circuit consists of a voltage source of $48 \, V$ connected in series with resistors of $100 \, \Omega$,$80 \, \Omega$,and $20 \, \Omega$. First,calculate the total resistance of the circuit: $R_{total} = 100 \, \Omega + 80 \, \Omega + 20 \, \Omega = 200 \, \Omega$. Next,calculate the total current $i$ flowing through the circuit using Ohm's law: $i = \frac{V}{R_{total}} = \frac{48 \, V}{200 \, \Omega} = 0.24 \, A$. The potential difference across $PQ$ corresponds to the voltage drop across the $20 \, \Omega$ resistor. Using Ohm's law for this resistor: $V_{PQ} = i 	imes R_{PQ} = 0.24 \, A 	imes 20 \, \Omega = 4.8 \, V$.

In the circuit shown,the potential difference across $PQ$ will be nearest to .............. $V$.

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