Question

In the network shown in figure, E; = 6V, E2 = 4V, R, = 212, R2 =392, R3 = 582. Find the currents passing through R1, R2 and R3.​

Answer 1

Answer:

Let us suppose that I  

1

​

 current flows through E  

1

​

 

and I  

2

​

 current flows through E  

2

​

 

Using Kirchhoff's Law:

I  

1

​

R  

1

​

+(I  

1

​

+I  

2

​

)R  

3

​

=E  

1

​

 

I  

1

​

×2+(I  

1

​

+I  

2

​

)5=17

2I  

1

​

+5I  

1

​

+5I  

2

​

=17

7I  

1

​

+5I  

2

​

=17

I  

1

​

R  

2

​

+(I  

1

​

+I  

2

​

)R  

3

​

=E  

2

​

 

I  

2

​

×3+(I  

1

​

+I  

2

​

)5=21

5I  

1

​

+8I  

2

​

=21

35I  

1

​

+25I  

2

​

=85

−

3

​

5I  

1

​

+  

−

5

​

6I  

2

​

=  

−

1

​

47

−31I  

2

​

=−62

​

 

I  

2

​

=2 amp

7I  

1

​

+5I  

2

​

=17

7I  

1

​

+10=17

7I  

1

​

=7

So, I  

1

​

=1 amp

Explanation: