Solve Current Electricity
Answers
hala mate,
Current flowing through various branches of the circuit is represented in the given figure.
Physics Class 12 Current Electricity Circuit 1
I1 = Current flowing through the outer circuit
I2 = Current flowing through branch AB
I3 = Current flowing through branch AD
I2 − I4 = Current flowing through branch BC
I3 + I4 = Current flowing through branch CD
I4 = Current flowing through branch BD
By Loop law:-
For the closed circuit ABDA, potential is zero i.e.
10I2 + 5I4 − 5I3 = 0
2I2 + I4 −I3 = 0
I3 = 2I2 + I4 … (1)
For the closed circuit BCDB, potential is zero i.e.
5(I2 − I4) − 10(I3 + I4) − 5I4 = 0
5I2 + 5I4 − 10I3 − 10I4 − 5I4 = 0
5I2 − 10I3 − 20I4 = 0
I2 = 2I3 + 4I4 … (2)
For the closed circuit ABCFEA, potential is zero i.e.
−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0
10 = 15I2 + 10I1 − 5I4
3I2 + 2I1 − I4 = 2 … (3)
From equations (1) and (2), we obtain
I3 = 2(2I3 + 4I4) + I4
I3 = 4I3 + 8I4 + I4
− 3I3 = 9I4
− 3I4 = + I3 … (4)
Putting equation (4) in equation (1), we obtain
I3 = 2I2 + I4
− 4I4 = 2I2
I2 = − 2I4 … (5)
It is evident from the given figure that,
I1 = I3 + I2 … (6)
Putting equation (6) in equation (1), we obtain
3I2 +2(I3 + I2) − I4 = 2
5I2 + 2I3 − I4 = 2 … (7)
Putting equations (4) and (5) in equation (7), we obtain
5(−2 I4) + 2(− 3 I4) − I4 = 2
− 10I4 − 6I4 − I4 = 2
17I4 = − 2
I4 = - (2/17)
Equation (4) reduces to
I3 = − 3(I4)
=-3(-2/17) = (6/17) A
I2=-2(I4)
=-2(-2/17) = (4/17) A
I2 - I4 = (4/17) – (-2/17) = (6/17) A
I3+ I4 = (6/17) + (-2/17) = (4/17) A
I1 = I3 + I2 = (6/17) + (4/17) = (10/17) A
Therefore, current in branch AB= (4/17) A
In branch BC = (6/17) A
In branch CD = (4/17) A
In branch AD= (6/17) A
In branch BD = (-2/17) A
Total Current =(4/17) +(6/17)+(-4/17)+(6/17)+(-2/17)=(10/17) A
hope it helps..