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Current Electricity question

2021 · 1 Sep · Shift 2 · Q55
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  5. /2021 · 1 Sep · Shift 2 · Q55

Current Electricity question

2021 · 1 Sep · Shift 2 · Q55

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
Two resistors R1 = (4 ±\pm± 0.8) Ω\OmegaΩ and R2 = (4 ±\pm± 0.4) Ω\OmegaΩ are connected in parallel. The equivalent resistance of their parallel combination will be :
  1. A
    (4 ±\pm± 0.4) Ω\OmegaΩ
  2. B
    (2 ±\pm± 0.4) Ω\OmegaΩ
  3. C
    (2 ±\pm± 0.3) Ω\OmegaΩ
  4. D
    (4 ±\pm± 0.3) Ω\OmegaΩ
View written solutionFree

Correct answer: C

  1. Equivalent resistance for parallel combination

For two resistors in parallel,

R=R1R2R1+R2R = \frac{R_1R_2}{R_1+R_2}R=R1​+R2​R1​R2​​

Given:

R1=(4±0.8) Ω,R2=(4±0.4) ΩR_1 = (4 \pm 0.8)\,\Omega, \qquad R_2 = (4 \pm 0.4)\,\OmegaR1​=(4±0.8)Ω,R2​=(4±0.4)Ω

Nominal value:

R=4×44+4=168=2 ΩR = \frac{4\times 4}{4+4} = \frac{16}{8} = 2\,\OmegaR=4+44×4​=816​=2Ω

So the equivalent resistance is centered at 2 Ω2\,\Omega2Ω.

  1. Find fractional error

Using

R=R1R2R1+R2R = \frac{R_1R_2}{R_1+R_2}R=R1​+R2​R1​R2​​

we take logarithmic/error form:

For product in numerator,

Δ(R1R2)R1R2=ΔR1R1+ΔR2R2\frac{\Delta (R_1R_2)}{R_1R_2} = \frac{\Delta R_1}{R_1} + \frac{\Delta R_2}{R_2}R1​R2​Δ(R1​R2​)​=R1​ΔR1​​+R2​ΔR2​​

For sum in denominator,

Δ(R1+R2)=ΔR1+ΔR2\Delta(R_1+R_2) = \Delta R_1 + \Delta R_2Δ(R1​+R2​)=ΔR1​+ΔR2​

so

Δ(R1+R2)R1+R2=0.8+0.44+4=1.28=0.15\frac{\Delta(R_1+R_2)}{R_1+R_2} = \frac{0.8+0.4}{4+4} = \frac{1.2}{8} = 0.15R1​+R2​Δ(R1​+R2​)​=4+40.8+0.4​=81.2​=0.15

Also,

ΔR1R1=0.84=0.2,ΔR2R2=0.44=0.1\frac{\Delta R_1}{R_1} = \frac{0.8}{4} = 0.2, \qquad \frac{\Delta R_2}{R_2} = \frac{0.4}{4} = 0.1R1​ΔR1​​=40.8​=0.2,R2​ΔR2​​=40.4​=0.1

Hence,

ΔRR=0.2+0.1+0.15=0.45\frac{\Delta R}{R} = 0.2 + 0.1 + 0.15 = 0.45RΔR​=0.2+0.1+0.15=0.45

This would give

ΔR=0.45×2=0.9 Ω\Delta R = 0.45\times 2 = 0.9\,\OmegaΔR=0.45×2=0.9Ω

which is clearly not among the options, and this direct addition is not the correct way here for a quotient involving a sum.

  1. Use differential method

Let

R=R1R2R1+R2R = \frac{R_1R_2}{R_1+R_2}R=R1​+R2​R1​R2​​

Then

ΔR≈∣∂R∂R1∣ΔR1+∣∂R∂R2∣ΔR2\Delta R \approx \left|\frac{\partial R}{\partial R_1}\right|\Delta R_1 + \left|\frac{\partial R}{\partial R_2}\right|\Delta R_2ΔR≈​∂R1​∂R​​ΔR1​+​∂R2​∂R​​ΔR2​

Now,

∂R∂R1=R22(R1+R2)2,∂R∂R2=R12(R1+R2)2\frac{\partial R}{\partial R_1} = \frac{R_2^2}{(R_1+R_2)^2}, \qquad \frac{\partial R}{\partial R_2} = \frac{R_1^2}{(R_1+R_2)^2}∂R1​∂R​=(R1​+R2​)2R22​​,∂R2​∂R​=(R1​+R2​)2R12​​

At R1=R2=4 ΩR_1=R_2=4\,\OmegaR1​=R2​=4Ω,

∂R∂R1=1664=14,∂R∂R2=1664=14\frac{\partial R}{\partial R_1} = \frac{16}{64} = \frac14, \qquad \frac{\partial R}{\partial R_2} = \frac{16}{64} = \frac14∂R1​∂R​=6416​=41​,∂R2​∂R​=6416​=41​

Therefore,

ΔR=14(0.8)+14(0.4)=0.2+0.1=0.3 Ω\Delta R = \frac14(0.8) + \frac14(0.4) = 0.2 + 0.1 = 0.3\,\OmegaΔR=41​(0.8)+41​(0.4)=0.2+0.1=0.3Ω
  1. Final equivalent resistance

Thus,

R=(2±0.3) ΩR = (2 \pm 0.3)\,\OmegaR=(2±0.3)Ω
  1. Check options
  • A: (4±0.4) Ω(4\pm0.4)\,\Omega(4±0.4)Ω ❌ nominal value wrong
  • B: (2±0.4) Ω(2\pm0.4)\,\Omega(2±0.4)Ω ❌ error wrong
  • C: (2±0.3) Ω(2\pm0.3)\,\Omega(2±0.3)Ω ✅
  • D: (4±0.3) Ω(4\pm0.3)\,\Omega(4±0.3)Ω ❌ nominal value wrong

So the correct option is C.

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