Question

a current in a series circuit is 15 A .when an additional 8 ohm resistor is inserted in series the current drops to 12 A what is the resistance in the original circuit

Answer 1

In initial condition,

Current in the circuit (i)=15A

Let the resistance be 'r' ohm initially.

Potential across the circuit=V

By the Ohm's Law;

V=ir

V=15r...................eq(1)

Now,a resistance of 8 ohm is connected in series.We know in series combination potential remains same as before in the circuit.So potential will be 'V' and current in the circuit (I) is 12A.So net resistance after connecting in series

R=r+8

By the Ohm's Law again,

V=I(r+8)

V=12(r+8)...............eq(2)

Now equating eq(1) and eq(2)

15r=12(r+8)

15r=12r+96

3r=96

r=32 ohm

Hence,in the original circuit resistance is 32ohm.