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A 100 Omega resistasnce is connected in ...

A `100 Omega` resistasnce is connected in series with a `4 H` inductor. The voltage across the resistor is `V_R=(2.0V)sin(10^3 rad//s)t`:
(a) Find the expession of circuit current
(b) Find the inductive reactance
(c) derive an expression for the voltage across the inductor,

Text Solution

Verified by Experts

a) Consider a series L-R circuit as shonw below

As `T=1/f = 1/60s`
`therefore omega=2pif=377 rads^(-1)`
`therefore X_(L) = omegaL=(377)(0.040)=15.80Omega`
`rArr Z=sqrt(X_(L)^(2)+R^(2))=25.05 Omega`
Phase single, `phi=tan^(-)(X_(L)/R)=tan^(-1) (0.754) =37^(@)`
b) Amplitude (maximum values) are,
`I_(0)= V_(0)/Z = 150/25.05=6A`
`rArr (V_(0))_(R) = I_(0)R=120 V` and `(V_(0))_(L)= I_(0)X_(L) = 90.5 V`
`V_(0) = sqrt((V_(0)_(R)^(2)+(V_(0))_(L)^(2))=sqrt((12)^(2)+(90.5)^(2))=15.03` V
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