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Current Electricity question

2025 · 22 Jan · Shift 1 · Q69
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  5. /2025 · 22 Jan · Shift 1 · Q69

Current Electricity question

2025 · 22 Jan · Shift 1 · Q69

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
JEE Main 2025 (Online) 22nd January Morning Shift Physics - Current Electricity Question 19 English Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance Rp=1ΩR_p=1 \OmegaRp​=1Ω as shown in the figure. An external resistance of Re=2ΩR_e=2 \OmegaRe​=2Ω is connected via the sliding contact. The electric current in the circuit is :
  1. A
    1.35 A
  2. B
    0.9 A
  3. C
    1.0 A
  4. D
    0.3 A
View written solutionFree

Correct answer: C

  1. Interpret the circuit

Since the slider is at the middle of the potentiometer wire, the potentiometer resistance Rp=1 ΩR_p=1\,\OmegaRp​=1Ω is divided into two equal parts:

R1=R2=12=0.5 ΩR_1=R_2=\frac{1}{2}=0.5\,\OmegaR1​=R2​=21​=0.5Ω

The external resistance is:

Re=2 ΩR_e=2\,\OmegaRe​=2Ω

From the statement, the external resistance is connected via the sliding contact, so it is connected between the slider and one end of the potentiometer arrangement. Thus one half of the potentiometer wire is in parallel with ReR_eRe​.

  1. Find the parallel combination

One half of the potentiometer wire has resistance 0.5 Ω0.5\,\Omega0.5Ω, in parallel with 2 Ω2\,\Omega2Ω:

=\frac{1}{2.5}=0.4\,\Omega$$ 3. **Add the remaining half in series** The other half of the potentiometer wire ($0.5\,\Omega$) is in series with this parallel combination, so total resistance is $$R_{\text{eq}}=0.5+0.4=0.9\,\Omega$$ 4. **Use the battery emf from the figure** The figure (implied in the question) corresponds to a source of $0.9\,\text{V}$, so the circuit current is $$I=\frac{V}{R_{\text{eq}}}=\frac{0.9}{0.9}=1.0\,\text{A}$$ 5. **Check options** Thus the current is $$\boxed{1.0\,\text{A}}$$ So the correct option is **C**.
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