SATHEE: current-electricity Question 25

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current-electricity Question 25

Question: Q. 18. A battery of emf $E$ and internal resistance, $r$ , when connected with an external resistance of $12 Ω$ produces a current of $0.5 A$ . When connected across a resistance of $25 Ω$ , it produces a current of 0.25 A. Determine (i) the emf and (ii) the internal resistance of the cell.

A [O.D. Comptt. I, II, III 2013]

Show Answer

Solution:

Ans. Given : $R_{1} = 12 Ω, I_{1} = 0.5 A$

From I condition,

$\begin{aligned} (i) & E & = I_{1} (R_{1} + r) E & = 0.5 (12 + r) & = 6 + 0.5 r \end{aligned}$

$O^{-} = 0.5 (12 + r)$

From II condition,

$\begin{aligned} (ii) & E & = I_{2} (R_{2} + r) E & = 0.25 (25 + r) & = 6.25 + 0.25 r \end{aligned}$

From eqns. (i) and (ii)

$\begin{aligned} 6 + 0.5 r & = 6.25 + 0.25 r \Rightarrow & 0.5 r - 0.25 r & = 6.25 - 6.0 \Rightarrow & 0.25 r & = 0.25 \Rightarrow & r & = 1 Ω Then & E & = 6 + 0.5 \times 1 E & = 6.5 V \end{aligned}$

$\Rightarrow r = 1 Ω 1 / 2$

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current-electricity Question 25

Table of Contents

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Test Platform

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