Question

Two cells of emf 1 V and 2 V and having respective internal resistances 2 ohm and 1 ohm are connected in parallel so as to send a current through an external resistance of 1 ohm in the same direction. Find the current through the cells and the current through the external resistance. ​

Answer 1

Let i1 and i2 be the current flowing through the two branches. the current through the 1 ohm resistor will be i1+i2(Kirchhoff's law)

Applying kirchoff's Voltage law to the loop ABCDEFA

-1(i1+i2)-i1+1=0

2i1+1i2=1. (1)

Applying kirchoff's law to loop BCDEB

-1(i1+i2)-2i2+1=0

1i1+3i2=1. (2)

Multiply equation (2) by 2

2i1+6i2=2. (3)

Substracting (1) and (3)

We get i2=1/5

Put i2=1/5 in (1)

2i1+1(1/5)=1

i1=1/10

Current through the 1ohm eternal resistance

i1+i2=1/5+1/10

=3/10 A