Let R be the resistance of the electric lamp.
In series total resistance =R1 + R2
=5 + R
1 = 10/5+R
R = 5 ohm
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
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
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