In the circuit given- E - 6-0 V- R1 - 100 - R2 - R3 - 50 - and R4 - 75 - The equivalent resistance of the circuit in - is

R_2, R_3, R_4 are in parllel. 1/R_eff amp;=1/R_2+1/R_3+1/R_4 amp;=1/50+1/50+1/75 R_e ff amp;=18.75 Ω For detailed solution watch the next video. Ne ...

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Byju's Answer

Standard XII

Physics

Series and Parallel Combination of Resistors

In the circui...

Question

In the circuit given, $E = 6.0 V, R_{1} = 100 Ω, R_{2} = R_{3} = 50 Ω$ and $R_{4} = 75 Ω$ . The equivalent resistance of the circuit $(in Ω)$ is

11.875

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26.31

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118.75

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None of these

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Solution

The correct option is C $118.75$
$R_{2}, R_{3}, R_{4}$ are in parllel.

$\begin{matrix} \frac{1}{R_{e f f}} & = \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}} = \frac{1}{50} + \frac{1}{50} + \frac{1}{75} R_{e f f} & = 18.75 Ω \end{matrix}$
For detailed solution watch the next video.

Net effective resistance is $118.75 Ω$

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Similar questions

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In the circuit given, E=6.0 V,R1=100 Ω,R2=R3=50 Ω and R4=75 Ω. The equivalent resistance of the circuit (in Ω) is

The correct option is C 118.75R2, R3, R4 are in parllel. 1Reff=1R2+1R3+1R4=150+150+175Reff=18.75 Ω For detailed solution watch the next video. Net effective resistance is 118.75 Ω

In the circuit given, $E = 6.0 V, R_{1} = 100 Ω, R_{2} = R_{3} = 50 Ω$ and $R_{4} = 75 Ω$ . The equivalent resistance of the circuit $(in Ω)$ is