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Question

# In the circuit shown in figure, E1=3V,E2=2V,E3=1V and r1=r2=r3=1ohm(a) Find the potential difference between the points A and B and the currents through each branch.(b) If r2 is short circuited and the point A is connected to point B through a resistance R, find the currents through E1,E2,E3 and the resistor R.

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Solution

## (a) Applying Kirchoff's loop law to mesh PLMQP and PLMQONP in the figure shown below, we havei1r1+i1r2=E1−E2ori2+i2=1.....(i)i1r1+i3r3=E1−E3ori1+i3=1.....(ii)At i2+i3=i1.....(iii)On solving (i), (ii), (iii)i1=1amp;i2=0amp;i3=1ampSince no current is drawn along the branck AP∴VAB=VPQPotential difference across PQVPQ=E1−i1r1=2volt(b) The figure shows the circuit when point A is connected to point B and r2 is short-circuited.Applying Kirchoff's junction rule at P, we geti=i1+i2+i3....(iv)Applying Kirchoff's law to mesh ABMLAi1r1=E1−E2;i1=1ampApplying Kirchoff's law to mesh ANOQMLi1r1+i3r3=E1−E3ori1−i3=2....(v)From above equationsi1=1amp,i2=2ampi3=−1amp (direction of current is opposite)So, current through resistor R will be I=I1+I2+I3=2amp

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