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# A cell of internal resistance r passes current for some time in a resistor R1 in another resistorR2 then show that r =underroot R1R2

Grade:12

## 1 Answers

Arun
25763 Points
one year ago
E = e.m.f of cell
r = internal resistance
P= Power = $i^{2}R$ = $\frac{V^{2}}{R}$ = $VR$
Current in circuit 1 and 2 is ${{i}__1}$ & ${{i}__2}$ respectively
CIRCUIT 1:

$E={{i}__1}r + {{i}__1}{{R}__1}$ __________(1)
$P={i_{1}^2}{{R}__1}$__________(2)
CIRCUIT 2:

$E={{i}__2}r + {{i}__2}{{R}__2}$ __________(3)
$P={i_{2}^2}{{R}__2}$__________(4)
From equation (1) & (3) we get
$r=({{{i}__2}{{R}__2}-{{i}__1}{{R}__1}})/({{{i}__1}-{{i}__2}})$__________(5)
From equation (2) & (4) we get
${{i}__1}^2/{{i}__2}^2={{R}__2}/{{R}__1}$__________(6)
From equation (5)
$=> r={{i}__2}[{{R}__2}-({{i}__1}/{{i}__2}){{R}__1}]\div {{i}__2}[({{i}__1}/{{i}__2}-1)]$
$({{i}__2} \ gets \ cancalled)$

Substituting equation (6)
$=>r=[{{R}__2}-{\sqrt{{}}{{{R}__1}{R}__2}}]\div [(\sqrt{{}}{{{R}__1}/{{R}__2}})-1]$
Symplifying and sending underroot R1 to the top and canceling (underroot R2  –  underroot R1) in numerator and denominator we finally get
$r=\sqrt{{{R}__1}{{R}__2}}$
$Ans: The \ internal \ resistance \ of \ the \ cell \ is \ \sqrt{{{R}__1}{{R}__2}}$

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