A current of 1 ampere flows in a series circuit having an electric lamp and a conductor of 5 ohm when connected to a 10 V battery. Calculate the resistance of the electric lamp.
Now if a resistance of 10 ohm is connected in parallel with this series combination, what change (if any) in current flowing through 5 ohm conductor and potential difference across the lamp will take place? Give reason.
OR
B1, B2, B3 are three identical bulbs connected as shown in the figure. Ammeters A1, A2, A3 are connected as shown. When all the bulbs glow, the current of 3A is recorded by ammeter A.
i. What happens to the glow of the other two bulbs when bulb B1 gets fused?
ii. What happens to the reading of A1, A2, A3 and A when the bulb B2 gets fused?
iii. How much power is dissipated in the circuit when all the three bulbs glow together?
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