QQuestion
Three identical bulbs B1, B2 and B3, each of power rating 18 W, 12 V, are connected to a battery of 12 V as shown.
(a) Calculate:
(i) the resistance of each bulb
(ii) the current drawn from the cell
(b) If bulb B3 is removed from the circuit, will the brightness of bulb B1 increase, decrease or remain the same?
✓Answer
Given,
Battery emf (ε) = 12 V
Voltage rating of each bulb (V) = 12 V
Power rating of each bulb (P) = 18 W
(a) 1. Let the resistance of each bulb be R.
Then, \mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}}
⇒ \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{12^2}{18}=\frac{144}{18}
⇒ R = 8 Ω
2. Since B2 and B3 are connected in parallel:
\frac{1}{\mathrm{R_P}}=\frac{1}{\mathrm{R_1}}+\frac{1}{\mathrm{R_2}}Here, R1 = R2 = 8 Ω
⇒ \frac{1}{\mathrm{R_P}}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}
⇒ RP = 4 Ω
This parallel combination is in series with B1.
Total resistance, RS = RP + R = 4 + 8 = 12 Ω
Therefore,
Current drawn = \frac{\text{Battery emf}}{\text{Total resistance}}=\frac{12}{12} = 1 A
Hence, the current drawn from the battery is 1 A.
(b) When B3 is removed, B1 and B2 remain in series.
Equivalent resistance = 8 + 8 = 16 Ω
Since the resistance increases, the current in the circuit decreases.
Hence, the brightness of B1 decreases.
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