Voltage at which it works, V = 6 V
As P = VI,
Resistance of the lamp,
If this lamp is connected to a source of 12 V supply directly, current through the lamp will be
= 12 /3 = 4A.
This current is double the current that the lamp can withstand. Thus, the purpose of the resistance (R) is to reduce the current to the safe value of 2 A.
Total resistance of the bulb and the resistance R = r + R. (in series)
For the efficient working of the bulb, the resistance R should be chosen in such a way that the resistance (r + R) should allow only a current of 2 A when connected in series with the 12 V supply, i.e.,
Or 6 + 2 R = 12
Or 2 R = 6
Or R = 3 Ω
Current in the circuit,
Since the resistances are in series, same current flows in each resistance.
Electric energy consumed by R1 (= 3 Ω) in 1 minute
W1 = I2R1 t = (4/3)2 × 3 × 60 = 320 J
Electric energy consumed by in R2 (= 6 Ω) = (4/3)2 × 6 × 60 = 640 J
(P = V2/R)
Resistance of the bulb B,
When the bulbs are connected in series, total resistance ,
R = R1 + R2 = 144 + 1440 = 1584 Ω
When these bulbs in series are connected across 120 V, current in the circuit, i.e.,
Power consumed by the bulb A, i.e., P1 = I2R1 = (0.076)2 × 144 = 0.832 W
Power consumed by the bulb B, i.e., P2 = I2R2 = (0.076)2 × 1440 = 8.32 W
Hence, the bulb B consumes more energy than the bulbs A, when they are connected in series.