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Two coil A and B have inductances 1.0 ...

Two coil A and B have inductances 1.0 H and 2.0 H respectively. The resistance of each coil is ` 10 Omega`. Each coil is connected to an ideal battery of emf `2.0 V` at `t = 0`. Let `i_A` and `i_B` be the current in the two circuit at time t. Find the raito `i_A/ I_B` at (a) `t=100ms`, (b) `t= 200 ms` and (c ) `t = 1 s`.

Text Solution

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Here `L_(A)=1.0H, L_(B)=2.0H,`
`R=10 (Omega)`
(a) `t=0.1s, T_(A)=0.1, T_(B)=(L/R)=0.2`
`i_(A)=i_(0)(1-e^(-t//(tau))),`
`=(2/10)(1-e^((0.1xx10)/1))=0.2(1-e^(-1))`
`=0.126424111`
`i_(B)=i_(0)(1-e^(-t//(tau))),`
`=(2/10)(1-e^(0.1xx10//2))`
`=0.2(1-e^(-1//2))=0.078693`
`(i_(A))/(i_(B))=(0.126411)/(0.78693)=1.6`
`(b) t=200ms=0.2s`
`l_(A)=0.2(1-e^(-0.2xx10.1))`
`=0.2xx0.864664716`
`=0.1729329943`
`l_(B)=0.2(1-e^(-0.2xx10.2))`
`=0.2xx0.632120=0.126424111`
`:. (i_A)/i_B)=(0.172932343)/(0.126424111)=1.36=1.4`
(c) `t=1s.`
`(i_A)=0.2(1-e^(-1xx10.1))`
`=0.2-0.9999546`
`=0.19999092`
`(i_B)=0.2(1-e^(-1xx10.2))`
`=0.2xx0.99326 = 0.19865241`
`:. (I_A)/(I_B)=(0.19999092)/(0.19999092)~~1.0`.
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