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Current Electricity question

2011 · Q165
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Current Electricity question

2011 · Q165

NEETPhysicsCurrent ElectricityMCQ+4 / −1
A current of 2 A flows through a 2 Ω\OmegaΩ resistor when connected across a battery a 2 Ω\OmegaΩ resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Ω\OmegaΩ resistor. The internal resistance of the battery is
  1. A
    0.5 Ω\OmegaΩ
  2. B
    1/3 Ω\OmegaΩ
  3. C
    1/4 Ω\OmegaΩ
  4. D
    1 Ω\OmegaΩ
View written solutionFree

Correct answer: B

Let ε\varepsilon ε be the emf and r be internal resistance of the battery.

In the first case,

AIPMT 2011 Prelims Physics - Current Electricity Question 84 English Explanation 1

2=ε2+r2 = {\varepsilon \over {2 + r}}2=2+rε​

In the second case,

AIPMT 2011 Prelims Physics - Current Electricity Question 84 English Explanation 2

0.5=ε9+r0.5 = {\varepsilon \over {9 + r}}0.5=9+rε​

Divide (i) by (ii), we get

20.5=9+r2+r⇒4+2r=4.5+0.5r{2 \over {0.5}} = {{9 + r} \over {2 + r}} \Rightarrow 4 + 2r = 4.5 + 0.5r0.52​=2+r9+r​⇒4+2r=4.5+0.5r

1.5r=0.5⇒r=0.51.5=13Ω1.5r = 0.5 \Rightarrow r = {{0.5} \over {1.5}} = {1 \over 3}\Omega 1.5r=0.5⇒r=1.50.5​=31​Ω

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