# Branch-current method

For a multi-loop circuit the following procedure is adopted for writing equations :

1. Assume currents in different branch of the network

2. Write down the smallest number of voltage drop loop equations so as to include all circuit elements ; these loop equations are independent.

If there are n nodes of three or more elements in a circuit, then write (n- 1) equations as per current law.

3. Solve the above equations simultaneously.

The assumption made about the directions of the currents initially is arbitrary. In case the actual direction is opposite to the assumed one, it will be reflected as a negative value for that current in the answer.

The branch-current method (the most primitive one) involves more labour and is not used except for very simple circuits.

Example 5. In the circuit of Fig. 21, find the current through each resistor and voltage drop across each resistor.

Solution. Let the currents be as shown in Fig. 22. Applying Kirchhoff’s voltage law to the circuit ABEF A, we get -12 I1 – 10 (I 1 + I 2 ) + 12 = 0

– 22 I1 + 16 I1 – 10 = 0

5 I1 + 8I2 – 5 = 0 Multiplying eqn. (i) by 5 and eqn. (ii) by 11 and subtracting, we get I2 = 0.397 A

Substituting this value in eqn. (i ), we get

11 I1 + 5 x 0.397-6 = 0

i.e.,            I1 = 0.365 A

Hence,             Current through 12 Q resistor, I1 = 0.365 A. (Ans.)

Current through 6 Q resistor,  I2 = 0.397 A. (Ans.)

Current through 10 Q resistor, I1+ 12 =0 .762 A. (Ans.)

The voltage drop across :

12 Q resistor = 0.365 x 12 = 4.38 V. (Ans.)

6 Q resistor = 0.397 x 6 = 2.38 V. (Ans.)

10 Q resistor = 0.762 x 10 = 7.62 V. (Ans.)

Example 6. Find the magnitude and direction of currents in each of the batteries shown in Fig. 23 Solution. Let the directions of currents h, 12 and 13 in the batteries be as shown in Fig. 24. Applying Kirchhoff’s voltage law to the circuit ABCFA, we get

30 – 4 I1 + 20 I2 – 20 – 12 I1 = 0 -24 I1 + 20 T1 + 10 = 0

12 – 20 I1 – 12 I1 + 10 = 0

6 I1 + 16 I2 – 15 = 0

Circuit ECDEF gives,

20 – 20 I1 ( I1 + I2) + 10 = 0

20 – 20 I2 – 12 I1 -12 I2 + 10 =0

– 12 I1 – 32 I2 + 30 = 0

I1 + 16 I2 – 15 = 0

Multiplying eqn. (ii) by 2 and subtracting it from (i), we get

– 42 I2 + 25 = 0

i.e.,            12 = 0.595 A

Substituting this value of ! 2 in eqn. (i), we get

12 I1 – 10 x 0.595-5 = 0

or     I1 = 0.912 A

Hence current through,

30 V battery,                           I1 = 0.912A. (Ans).

20 V battery,                           12 = 0.595 A. (Ans.)

10 V battery,                           I1 + 12) = 1.507 A. (Ans).
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