Question 14: Current Electricity

Question

In the given circuit, $ E_{1}=E_{2}=E_{3}=2 \mathrm{~V} $ and $ R_{1}=R_{2}=4 \Omega $, then current flowing through the branch $ A B $ is
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m7m77bl9/9fb65970-2735-4920-815b-058cc682404f/550f31d0-f468-11ef-bbe5-33f4ad49c5b5/file-jaoe38c1m7m77bla.png?format=png)

Accepted Answer

The given circuit can be redrawn as,
where, equivalent emf of the combination,
$ \begin{aligned} E_{\mathrm{eq}} & =\frac{E_{1} R_{1}+E_{2} R_{2}}{R_{1}+R_{2}} \ & =\frac{2 	imes 4+2 	imes 4}{4+4}=2 \mathrm{~V} \end{aligned} $
and $ R_{e q}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{4 	imes 4}{4+4}=\frac{16}{8}=2 \Omega $
Net emf of the circuit,
$ \begin{aligned} E^{\prime} & =E_{2}+E_{\mathrm{eq}} \ & =2+2 \ & =4 \mathrm{~V} \end{aligned} $
$ 	herefore $ Current flowing through branch $ A B $,
$ I=\frac{E^{\prime}}{R_{\mathrm{eq}}}=\frac{4}{2}=2 \mathrm{~A} $
Clearly from figure, it is clear that direction of current is from point $ A $ to $ B $ clockwise is loop.
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m7m7mq7j/218feea9-1393-4a7d-b5b9-2271b428aaf4/017d3bf0-f46a-11ef-bbe5-33f4ad49c5b5/file-jaoe38c1m7m7mq7k.png?format=png)

Answer

zero

Answer

$ 2 A $ from $ A $ and $ B $

Answer

$ 2 A $ from $ B $ to $ A $

Answer

$ 5 A $ from $ B $ to $ B $

Current Electricity

In the given circuit, E_{1}=E_{2}=E_{3}=2 \mathrm{~V} and R_{1}=R_{2}=4 \Omega , then current flowing through the branch A B is

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