In an electrical circuit four identical bulbs are connected in parallel to each other to a battery of 10 V (negligible internal resistance). When all the four bulbs glow, a current of 4 A is recorded. Then the power dissipated in the circuit and the resistance of each bulb are ________ and ________, respectively.  

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RRB Group D 19 Sept 2022 Shift 1 Official Paper
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  1. 50 W, 15 Ω 
  2. 30 W, 15 Ω
  3. 25 W, 40 Ω
  4. 40 W, 10 Ω

Answer (Detailed Solution Below)

Option 4 : 40 W, 10 Ω
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Detailed Solution

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Given: 

Four identical bulbs are connected parallelly.

The voltage of the battery = 10 V

Current = 4 A 

Formula used:

Power = V × I

From Ohm's Law, V = I × R

Effective resistance in parallel, \({1\over R_ {eq}} = {1\over R_ {1}} +{1\over R_ {2}}+ {1\over R_ {3}} +{1\over R_ {4}} \)

Calculation:

\(V = I × R_{eq}\\ R_{eq}= {10 \over 4}\)

Resistance in each bulb(R),

 \({1\over R_ {eq}} = {1\over R_ {1}} +{1\over R_ {2}}+ {1\over R_ {3}} +{1\over R_ {4}} \\ ={1\over R} +{1\over R}+ {1\over R} +{1\over R} \\ ={4\over R}\\ R_{eq}= {R \over 4}\\ R = {10\over 4} \times 4\\ R=10 \:\Omega\)

Thus, resistance in each bulb is 10 Ω. 

Current in each bulb, 

\(I = {V \over R} \\ I={10 \over 10}\\I=1\:A\)

Power dissipated in the circuit, 

\(P=V × I\\ =10 \times 4 \\=40\)

 
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