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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

The correct Answer is:
A, B, C, D
(a) `I_0=((V_0)_R)/R=2.5/300`
`=8.33xx10^-3A`
`=8.33mA`
Curent function and `V_R` function are in phase.
Hence,
`I(8.33mA)cos[(950rad//s)t]`
(b) `X_L=omegaL=950xx0.8=760Omega`
(c) `(V_0)_L=I_0X_L`
`=(8.33xx10^-3)(760)`
`=6.33V`
Now `V_L` function leads the current (or `V_R`) function by `90^@`
`:. V_L=6.33cos(950t+90^@)`
`=-6.33sin(950 t)`.
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