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An LR circuit constains an inductor of ...

An LR circuit constains an inductor of 500 mH, a resistor of `25.O Omega` and an emf of 5.00 V in series. Find the potential difference across the resistor at t =(a) 20.0 ms, (b) 100 ms and ( c) 1.00 s.

Text Solution

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Here `L=500mH, R=25 Omega, E=5V`
`(a) For t=20 ms`
` i=i_0(1-e^(-tR//L)`
`=(E/R)(1-e^(-tR//L))`
`=(5/25)(1-e^(-(2xx10^(-3)xx25)//(500xx10^(-3))`
`=1/5(1-e^(-1)=(1/5)(1-0.3678)`
`=(0.632)/(5)=0.1264`.`
`Potential difference `
`=iR=C.1264xx25`
`=3.1606V=3.16V`
`(b) Here t=100ms`
`i-i_0(1-e^(-tR//L))`
`=(5/25)(1-r^(-100)xx10^(-3)xx25//500xx10^(-3))`
`=1/5(1-e^(-50)`
=(1/5)(1-0.0067)`
`=(0.9932)/5=0.19864`
potential difference iR
`0.19864xx25xx4.9665=4.97 V.`
` (c ) Here t=1 sec.`
`i=(5/25)(1-e^(-1xx(25/100)xx10^(-3))`
`=(1/5)(1-e^(-50))`
`=1/5xx1=(1/5)A`
potential difference `
`=(1/5 xx25)V=5V.`
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