Question 720: Current Electricity

Question

Two ideal batteries of emf V
1
and V
2
and three resistances R
1
, R
2
and R
3
are connected as shown in the figure. The current in resistance R
2
would be zero if
![](https://app-content.cdn.examgoal.net/fly/@width/image/1ky4iki7j/7a879e3a-b2cf-404b-84ee-14b0e303156b/f05f4300-6fc8-11ec-b60f-27942bef1403/file-1ky4iki7k.png?format=png)

Accepted Answer

Let i
1
and i
2
be the currents as shown in the figure.
Apply Kirchhoff's law in the loop ABCDA and CEFDC to get
${i_1}{R_1} + ({i_1} - {i_2}){R_2} = {V_1}$ ..... (1)
${i_2}{R_3} - ({i_1} - {i_2}){R_2} = {V_2}$ ...... (2)
Multiply equation (1) by R
3
and (2) by R
1
and then subtract to get the current through R
2
as
$({i_1} - {i_2}) = {{{V_1}{R_3} - {V_2}{R_1}} \over {{R_1}{R_3} + {R_2}{R_3} - {R_1}{R_2}}}$.
The current through R
2
becomes zero when V
1
R
3
= V
2
R
1
.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1ky4imiet/765770d6-1a05-49c9-a335-8558d5a308af/28258560-6fc9-11ec-8887-d75613c1bf3a/file-1ky4imieu.png?format=png)

Answer

V
1
= V
2
and R
1
= R
2
= R
3

Answer

V
1
= V
2
and R
1
= 2R
2
= R
3

Answer

V
1
= 2V
2
and 2R
1
= 2R
2
= R
3

Answer

2V
1
= V
2
and 2R
1
= R
2
= R
3

Current Electricity

Two ideal batteries of emf V 1 and V 2 and three resistances R 1 , R 2 and R 3 are connected as shown in the figure. The current in resistance R 2 would be zero if

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