Question

In the circuit shown in figure E1 = 3 volts, E2 = 2 volts, Ez = 1 volt and R = r1 = r2 = 13 = 1 ohm.

(a)Find the potential difference between the points A and B and the currents through each branch.​

Answer 1

Answer:

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Explanation:

ANSWER

(a) Applying Kirchoff's loop law to mesh PLMQP and PLMQONP in the figure shown below, we have

i

1

r

1

+i

1

r

2

=E

1

−E

2

ori

2

+i

2

=1.....(i)

i

1

r

1

+i

3

r

3

=E

1

−E

3

ori

1

+i

3

=1.....(ii)

At i

2

+i

3

=i

1

.....(iii)

On solving (i), (ii), (iii)

i

1

=1amp;i

2

=0amp;i

3

=1amp

Since no current is drawn along the branck AP

∴V

AB

=V

PQ

Potential difference across PQ

V

PQ

=E

1

−i

1

r

1

=2volt

(b) The figure shows the circuit when point A is connected to point B and r

2

is short-circuited.

Applying Kirchoff's junction rule at P, we get

i=i

1

+i

2

+i

3

....(iv)

Applying Kirchoff's law to mesh ABMLA

i

1

r

1

=E

1

−E

2

;i

1

=1amp

Applying Kirchoff's law to mesh ANOQML

i

1

r

1

+i

3

r

3

=E

1

−E

3

ori

1

−i

3

=2....(v)

From above equations

i

1

=1amp,

i

2

=2amp

i

3

=−1amp (direction of current is opposite)

So, current through resistor R will be I=I

1

+I

2

+I

3

=2amp