Question

a current of 1 ampere flows in a series circuit containing a electric lamp and a conductor.... full questions in image​

Answer 1

Answer:

We know that, Effective resistance = V/I = 10 ohms

Hence, in series

R1 + R2 = 10

5 + R2 =10

Hence R2 = 5 ohms

The resistance of the electric lamp is 5 ohms

In a parallel circuit, the current changes, but voltage remains the same.

So, in parallel

1/R1 + 1/R2 = 1/R eff

= 2/10

R eff = 5 ohms

Hence, if we find out the value of the current = V/R = I

= 10/5 = 2

Hence the current set in the conductor changed from 1 A to 2 A if the setup is in parallel.

The circuit diagram is given above.

Hope this helps!!