Q.
The value of the current I1,I2 and I3 flowing through the circuit given below is
1842
209
NTA AbhyasNTA Abhyas 2020Current Electricity
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Solution:
Applying Kirchoff's law 10−6IB+14−4IA=0 ...(loop 1) 24=6IB+4IA 12=3IB+2IA .....(i) IA=IB+IC .....(junction) 10−6IB+2IC=0 .....(loop 2) 10=6IB−2IC 10=6IB−2(IA−IB) .....(ii)
On solving Equations (i) and (ii), we get IB=2A,IA=3A,IC=1A
Now as per given circuit, I1=IB=2A,I2=−IA=−3A I3=−IC=−1A