Question 321: Current Electricity

Question

A cell E
1
of emf 6V and internal resistance 2$\Omega$ is connected with another cell E
2
of emf 4V and internal resistance 8$\Omega$ (as shown in the figure). The potential difference across points X and Y is :
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l9fny6ll/5a43dc9a-3ed7-45ad-a19b-de96761a6526/fab04d90-4fb0-11ed-88d9-abb495ae8a3e/file-1l9fny6lm.png?format=png)

Accepted Answer

The circuit can be shown as below
The current through the circuit,
$I = {{{\varepsilon _1} - {\varepsilon _2}} \over {{r_1} + {r_2}}} = {{6 - 4} \over {10}} = {1 \over 5}A$
$	herefore$ Potential difference across points X and Y is
${V_{XY}} = {E_2} + I{r_2} = 4 + {1 \over 5} 	imes 8 = 5.6V$
![](https://app-content.cdn.examgoal.net/fly/@width/image/1kxemxwjv/abf9ed61-3b59-475a-bf8c-6324b7a2f67e/a899d6a0-618d-11ec-b27d-5f9767b4f07e/file-1kxemxwjw.png?format=png)

Answer

10.0 V

Answer

2.0 V

Answer

5.6 V

Answer

3.6 V

Current Electricity

A cell E 1 of emf 6V and internal resistance 2\Omega is connected with another cell E 2 of emf 4V and internal resistance 8\Omega (as shown in the figure). The potential difference across points X and Y is :

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