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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

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
25750 Points
4 years 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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