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Current Electricity question

2012 · Q81
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Current Electricity question

2012 · Q81

NEETPhysicsCurrent ElectricityMCQ+4 / −1
A cell having an emf ε\varepsilonε and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by
  1. A
    AIPMT 2012 Mains Physics - Current Electricity Question 86 English Option 1
  2. B
    AIPMT 2012 Mains Physics - Current Electricity Question 86 English Option 2
  3. C
    AIPMT 2012 Mains Physics - Current Electricity Question 86 English Option 3
  4. D
    AIPMT 2012 Mains Physics - Current Electricity Question 86 English Option 4
View written solutionFree

Correct answer: C

AIPMT 2012 Mains Physics - Current Electricity Question 86 English Explanation

Current in the circuit, I=εR+rI = {\varepsilon \over {R + r}}I=R+rε​

Potential difference across R,

V=IR=(εR+r)RV = IR = \left( {{\varepsilon \over {R + r}}} \right)RV=IR=(R+rε​)R

V=ε1+rRV = {\varepsilon \over {1 + {r \over R}}}V=1+Rr​ε​

When R = 0, V = 0 and when R = ∞\infty ∞, V = ε\varepsilon ε

∴\therefore∴ option (c) represents the correct graph.

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