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Current Electricity question

2022 · Q183
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Current Electricity question

2022 · Q183

NEETPhysicsCurrent ElectricityMCQ+4 / −1

The sliding contact C is at one fourth of the length of the potentiometer wire (AB) from A as shown in the circuit diagram. If the resistance of the wire AB is R0, then the potential drop (V) across the resistor R is

NEET 2022 Phase 2 Physics - Current Electricity Question 19 English

  1. A
    2V0R2R0+3R{{2{V_0}R} \over {2{R_0} + 3R}}2R0​+3R2V0​R​
  2. B
    4V0R3R0+16R{{4{V_0}R} \over {3{R_0} + 16R}}3R0​+16R4V0​R​
  3. C
    4V0R3R0+R{{4{V_0}R} \over {3{R_0} + R}}3R0​+R4V0​R​
  4. D
    2V0R4R0+R{{2{V_0}R} \over {4{R_0} + R}}4R0​+R2V0​R​
View written solutionFree

Correct answer: B

NEET 2022 Phase 2 Physics - Current Electricity Question 19 English Explanation

Equivalent resistance across point AC

$${R_{AC}} = {{{{{R_0}} \over 4} \times R} \over {{{{R_0}} \over 4} + R}} = {{R{R_0}} \over {{R_0} + 4R}}$$

From voltage divider rule

$${V_{AC}} = {{{R_{AC}}} \over {{R_{AC}} + {R_{CB}}}}{V_0} = {{{{R{R_0}} \over {{R_0} + 4R}}{V_0}} \over {{{R{R_0}} \over {{R_0} + 4R}} + {{3{R_0}} \over 4}}}$$

$$ = {{4R{R_0}{V_0}} \over {4R{R_0} + 3R_0^2 + 12R{R_0}}} = {{4R{V_0}} \over {3{R_0} + 16R}}$$

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