In the circuit given, E = 6.0 , V, R_1 = 100 | vedclass

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(C) In the given circuit, the resistors $R_2$, $R_4$, and $R_3$ are connected in parallel.The equivalent resistance $R_p$ of these three resistors is

Vedclass · Accepted Answer

$118.75$

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(C) In the given circuit, the resistors $R_2$, $R_4$, and $R_3$ are connected in parallel.The equivalent resistance $R_p$ of these three resistors is given by:$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_4} + \frac{1}{R_3}$Substituting the given values:$\frac{1}{R_p} = \frac{1}{50} + \frac{1}{75} + \frac{1}{50}$$\frac{1}{R_p} = \frac{3 + 2 + 3}{150} = \frac{8}{150}$$R_p = \frac{150}{8} = 18.75 \, \Omega$This equivalent resistance $R_p$ is in series with resistor $R_1$.Therefore, the total equivalent resistance $R_{eq}$ of the circuit is:$R_{eq} = R_1 + R_p$$R_{eq} = 100 + 18.75 = 118.75 \, \Omega$

In the circuit given, $E = 6.0 \, V$, $R_1 = 100 \, \Omega$, $R_2 = R_3 = 50 \, \Omega$, and $R_4 = 75 \, \Omega$. The equivalent resistance of the circuit, in $\Omega$, is:

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