**Question:**

Determine the current in each branch of the network shown in fig 3.30:

**Solution:**

Current flowing through various branches of the circuit is represented in the given figure.

*I*1 = Current flowing through the outer circuit

*I*2 = Current flowing through branch AB

*I*3 = Current flowing through branch AD

*I*2 − *I*4 = Current flowing through branch BC

*I*3 + *I*4 = Current flowing through branch CD

*I*4 = Current flowing through branch BD

For the closed circuit ABDA, potential is zero i.e.,

$10 l_{2}+5 l_{4}-5 l_{3}=0$

$2 I_{2}+I_{4}-I_{3}=0$

$I_{3}=2 I_{2}+I_{4} \ldots$ (1)

For the closed circuit BCDB, potential is zero i.e.,

5(*I*2 − *I*4) − 10(*I*3 +* I*4) − 5*I*4 = 0

5*I*2 + 5*I*4 − 10*I*3 − 10*I*4 − 5*I*4 = 0

5*I*2 − 10*I*3 − 20*I*4 = 0

*I*2 = 2*I*3 + 4*I*4 … (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (*I*1) + 10(*I*2) + 5(*I*2 −* I*4) = 0

10 = 15*I*2 + 10*I*1 − 5*I*4

3*I*2 + 2*I*1 − *I*4 = 2 … (3)

From equations (1) and (2), we obtain

*I*3 = 2(2*I*3 + 4*I*4) +* I*4

*I*3 = 4*I*3 + 8*I*4 +* I*4

− 3*I*3 = 9*I*4

− 3*I*4 = + *I*3 … (4)

Putting equation (4) in equation (1), we obtain

*I*3 = 2*I*2 + *I*4

− 4*I*4 = 2*I*2

*I*2 = − 2*I*4 … (5)

It is evident from the given figure that,

*I*1 = *I*3 + *I*2 … (6)

Putting equation (6) in equation (1), we obtain

3*I*2 +2(*I*3 + *I*2) −* I*4 = 2

5*I*2 + 2*I*3 − *I*4 = 2 … (7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2* I*4) + 2(− 3 *I*4) − *I*4 = 2

− 10*I*4 − 6*I*4 − *I*4 = 2

17*I*4 = − 2

$I_{4}=\frac{-2}{17} \mathrm{~A}$

Equation (4) reduces to

$I_{3}=-3\left(I_{4}\right)$

$=-3\left(\frac{-2}{17}\right)=\frac{6}{17} \mathrm{~A}$

$I_{2}=-2\left(I_{4}\right)$

$=-2\left(\frac{-2}{17}\right)=\frac{4}{17} \mathrm{~A}$

$I_{2}-I_{4}=\frac{4}{17}-\left(\frac{-2}{17}\right)=\frac{6}{17} \mathrm{~A}$

$I_{3}+I_{4}=\frac{6}{17}+\left(\frac{-2}{17}\right)=\frac{4}{17} \mathrm{~A}$

$I_{1}=I_{3}+I_{2}$

$=\frac{6}{17}+\frac{4}{17}=\frac{10}{17} \mathrm{~A}$

Therefore, current in branch $\mathrm{AB}=\frac{4}{17} \mathrm{~A}$

In branch $B C=\frac{6}{17} A$

In branch $C D=\frac{-4}{17} A$

In branch $A D=\frac{6}{17} A$

In branch $B D=\left(\frac{-2}{17}\right) A$

Total current $=\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17} \mathrm{~A}$

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.