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Current Electricity question

2006 · Q134
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Current Electricity question

2006 · Q134

NEETPhysicsCurrent ElectricityMCQ+4 / −1
Power dissipated across the 8 Ω\OmegaΩ resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the 3 Ω\OmegaΩ resistor is

AIPMT 2006 Physics - Current Electricity Question 65 English
  1. A
    3.0
  2. B
    2.0
  3. C
    1.0
  4. D
    0.5
View written solutionFree

Correct answer: A

Power = V . I = I2R

i2=PowerR=28{i_2} = \sqrt {{{Power} \over R}} = \sqrt {{2 \over 8}} i2​=RPower​​=82​​

= 14=12A\sqrt {{1 \over 4}} = {1 \over 2}A41​​=21​A

Potential over 8Ω=Ri2=8×12=4V8\Omega = R{i_2} = 8 \times {1 \over 2} = 4V8Ω=Ri2​=8×21​=4V

This is the potential over parallel branch. So,

i1=44=1A{i_1} = {4 \over 4} = 1Ai1​=44​=1A

Power of 3Ω=i12R=1×1×3=3W3\Omega = {i_1}^2R = 1 \times 1 \times 3 = 3W3Ω=i1​2R=1×1×3=3W

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