A current of 1 am

Part 1:

Let R be the resistance of the electric lamp.

In series total resistance =R₁ + R₂

=5 + R

1 = 10/5+R

R = 5 ohm

Part 2:

Total Resistance =

= = 100/20= 5ohm

Current in each branch= = 1 Amp (since both branches have same resistances, current divides equally)

V across Lamp + conductor = 10 V

V across Lamp = I × R = 1 × 5 = 5 Volt

OR

1. When bulb B₁ gets fused, no current will pass through it but the potential difference across the other two bulbs will remain same and hence, the glow of the bulbs B₂ and B₃ will remain the same.

2. When all three bulbs glow the reading in ammeter A= 3A, which means the current is divided equally amongst all the three branches of the circuit. Thus, A₁=A₂=A₃= 1A

When bulb B₂ gets fused no current will pass through it so the reading in ammeter A₂ is 0A.

Reading on A1 is 1 ampere, reading on A2 is 0 ampere, reading on A3 is 1 ampere and reading on A is A1+A2= 2 ampere.

3. Given: Voltage in the circuit= 4.5V

Total current in the circuit= 3A

To calculate: Power dissipated in the circuit

P=V.I=4.5× 3=13.5 W

Power dissipated in the circuit when all the three bulbs glow together= 13.5W