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Current Electricity question

2013 · Q124
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Current Electricity question

2013 · Q124

NEETPhysicsCurrent ElectricityMCQ+4 / −1
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω\OmegaΩ is
  1. A
    0.8 Ω\OmegaΩ
  2. B
    1.0 Ω\OmegaΩ
  3. C
    0.2 Ω\OmegaΩ
  4. D
    0.5 Ω\OmegaΩ
View written solutionFree

Correct answer: D

NEET 2013 Physics - Current Electricity Question 94 English Explanation

I=εR+rI = {\varepsilon \over {R + r}}I=R+rε​

⇒IR+Ir=ε\Rightarrow IR + Ir = \varepsilon⇒IR+Ir=ε

Here,
R=10Ω,r=?,ε=2.1V,I=0.2AR = 10\Omega ,r = ?,\varepsilon = 2.1V,I = 0.2AR=10Ω,r=?,ε=2.1V,I=0.2A

∴\therefore∴ 0.2 × 10 + 0.2 × r = 2.1

2 + 0.2r = 2.1

0.2r = 0.1 ⇒\Rightarrow⇒ r = 12{1 \over 2}21​ = 0.5 Ω\Omega Ω

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