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Current Electricity question

2016 · 9 Apr · Shift 1 · Q70
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Current Electricity question

2016 · 9 Apr · Shift 1 · Q70

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
JEE Main 2016 (Online) 9th April Morning Slot Physics - Current Electricity Question 286 English In the circuit shown, the resistance r is a variable resistance. If for r = fR, the heat generation in r is maximum then the value of f is :
  1. A
    14{1 \over 4}41​
  2. B
    12{1 \over 2}21​
  3. C
    34{3 \over 4}43​
  4. D
    1
View written solutionFree

Correct answer: B

  1. Use maximum power transfer idea

    Heat generated in the variable resistor rrr is Pr=I2rP_r = I^2 rPr​=I2r and for a linear resistive network, this power is maximum when r=Rthr = R_{\text{th}}r=Rth​ where RthR_{\text{th}}Rth​ is the Thevenin resistance of the remaining circuit seen from the terminals of rrr.

  2. Find the equivalent resistance seen by rrr

    From the given circuit, when we remove rrr and look into its terminals, the remaining network reduces to an equivalent resistance of Rth=R2.R_{\text{th}} = \frac{R}{2}.Rth​=2R​.

  3. Condition for maximum heat in rrr

    Therefore, for maximum heat generation, r=Rth=R2.r = R_{\text{th}} = \frac{R}{2}.r=Rth​=2R​.

  4. Compare with r=fRr = fRr=fR

    Given r=fR,r = fR,r=fR, so fR=R2  ⟹  f=12.fR = \frac{R}{2} \implies f = \frac{1}{2}.fR=2R​⟹f=21​.

  5. Check options

    f=12f = \frac{1}{2}f=21​ corresponds to Option B.

Final Answer: 12\boxed{\frac{1}{2}}21​​

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