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Question

Three identical bulbs B1, B2, and B3 each of power rating 18 W, 12 V are connected to a battery of 12 V.
- Calculate:
- the resistance of each bulb
- the current drawn from the cell
- If the bulb B3 is removed from the circuit, then will the brightness of the bulb B1 increase, decrease or remain the same?
Long Answer
Numerical
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Solution
a.
1. Power = `("Voltage")^2/"Resistance"`
Resistance = `("Voltage")^2/"Power"`
= `(12 xx 12)/18`
= `144/18`
= 8 ohm
2. `1/R_P = 1/R_2 + 1/R_3`
RP = `(R_2 xx R_3)/(R_2 + R_3)`
= `(8 xx 8)/(8 + 8)`
= `64/16`
= 4 ohm
Total resistance (RT) = 8 + 4 = 12 Ω
∴ The current drawn from the cell:
I = `V/R_T`
= `12/12`
= 1 A
b. If B3 is removed, then total resistance = 8 + 8 = 16 W
The brightness of bulb B decreases due to less current in the circuit.
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