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ALLEN 31-8-14 QNo. 29

ALLEN 31-8-14 QNo. 29

Question Image
Grade:12th pass

1 Answers

Shivam Chopra
45 Points
6 years ago
The answer is 1Ω
Here as the bulb is glowing at full intensity, so voltage across the bulb is = 1.5V
Also, for bulb, v = 1.5V, P = 0.45W;
Resistance = R = v2/P = 1.5*1.5/0.45 = 5Ω
So, total resistance b/w X & Y is R = (5*R) / (R+5) Ω
Total resistance of circuit = ((5*R) / (R+5)) + 3Ω
                                              = (8R + 15) / (R+5)Ω
Also, voltage across 3Ω resistance = 6-1.5 = 4.5V
 
=> 3 / { (8R + 15) / (R + 5) } * 6 = 4.5V
(Applying  formula {( Given resistance / Total resistance ) * Total voltage = Voltage on given resistance} only applicable for series combination of resistances )
The value of R comes to be = 1.25Ω
So, total resistance = (5R) / (R+5) =( 5*1.25) / (5+1.25) = 1Ω
Please approve it if you like.
Thanks
The answer is 1Ω
Here as the bulb is glowing at full intensity, so voltage across the bulb is = 1.5V
Also, for bulb, v = 1.5V, P = 0.45W;
Resistance = R = v2/P = 1.5*1.5/0.45 = 5Ω
So, total resistance b/w X & Y is R = (5*R) / (R+5) Ω
Total resistance of circuit = ((5*R) / (R+5)) + 3Ω
                                              = (8R + 15) / (R+5)Ω
Also, voltage across 3Ω resistance = 6-1.5 = 4.5V
 
=> 3 / { (8R + 15) / (R + 5) } * 6 = 4.5V
(Applying  formula {( Given resistance / Total resistance ) * Total voltage = Voltage on given resistance} only applicable for series combination of resistances )
The value of R comes to be = 1.25Ω
So, total resistance = (5R) / (R+5) =( 5*1.25) / (5+1.25) = 1Ω
Please approve it if you like.
Thanks

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