QQuestion
If two electric bulbs of ratings ‘240V-60 W’ and ‘240V-100W’ are connected in series, which one will shine brighter? (Filaments of both the bulbs are of the same material).[3]
OR
An electric cell of internal resistance 5 Ω and emf 2 V is connected to a resistor of 15 Ω. Determine the potential difference between the two ends of the cell. [3]
✓Answer
Find resistance of each (R = V²/P):
R₆₀ = 240² / 60
= 57600 / 60
= 960 Ω
R₁₀₀ = 240² / 100
= 57600 / 100
= 576 Ω
Series current:
R_total = 960 + 576 = 1536 Ω
I = V / Rtotal = 240 / 1536 = 0.15625 A
Power in each (P = I²R):
P₆₀ = (0.15625)² × 960 = 23.44 W
P₁₀₀ = (0.15625)² × 576 = 14.06 W
Conclusion: The 60 W bulb (higher resistance) dissipates more power in series and shines brighter.
Thus, the 60W bulb will shine brighter than the 100W bulb in this series connection.
OR
EMF = electromotive force of the cell = 2 V
r = internal resistance of the cell = 5 Ω
R = external resistor = 15 Ω
RTotal = R + r = 15 + 5 = 20 Ω
Current (I) = 2\over 20 = 0.1 A
V = EMF − I⋅r = 2V – (0.1 A × 5 Ω) = 1.5 V
Thus, the potential difference between the two ends of the cell is 1.5 V.
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