In the circuit given E=0.6V, $R_{1}$=100$\Omega$, $R_{2} = R_{3} = 50\Omega,\ R_{4} = 75\Omega$. The equivalent resistance of the circuit, in ohm is

Question

Dr. Abubacker · Accepted Answer

(c) $R_{2},\ R_{3}and\ R_{4}$ are in parallel order, so their equivalent resistance
$\frac{1}{R^{'}} = \frac{1}{R^{2}} + \frac{1}{R^{3}} + \frac{1}{R^{4}}$
$\ \ \ \ \ \  = \frac{1}{50} + \frac{1}{50} + \frac{1}{75}$
$\ \ \ \ \ \  = \frac{30 + 30 + 20}{1500}$
$\ \ \ \ \ \  = \frac{80}{1500} = \frac{4}{75}\ $
$	herefore\ \ R^{'} = \frac{75}{4}\Omega$
$R = R_{1} + R^{'} = 100 + \frac{75}{4}$
$\ \ \ \  = \frac{475}{4}\Omega = 118.75\Omega$

Q. In the circuit given E=0.6V, $R_{1}$=100$\Omega$, $R_{2} = R_{3} = 50\Omega,\ R_{4} = 75\Omega$. The equivalent resistance of the circuit, in ohm is

Explanation

Reviewed by Dr. Abubacker

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