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(B) The circuit is a voltage divider with two resistors in series, $R_1 = 64 \, \Omega$ and $R_2 = 32 \, \Omega$, connected across a potential differe

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The circuit is a voltage divider with two resistors in series, $R_1 = 64 \, \Omega$ and $R_2 = 32 \, \Omega$, connected across a potential difference of $60 	ext{ V} - 0 	ext{ V} = 60 	ext{ V}$.
The total resistance of the circuit is $R_{eq} = R_1 + R_2 = 64 \, \Omega + 32 \, \Omega = 96 \, \Omega$.
The current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{60 	ext{ V}}{96 \, \Omega} = \frac{5}{8} 	ext{ A}$.
The potential difference between points $P$ and $Q$ is the voltage across the $32 \, \Omega$ resistor.
$V_{PQ} = I \cdot R_2 = \left( \frac{5}{8} 	ext{ A} ight) 	imes 32 \, \Omega = 5 	imes 4 = 20 	ext{ V}$.

In the given circuit, the potential difference between points $P$ and $Q$ is . . . . . . (in $\text{ V}$)

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