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A 100 Omega resistance is connected in s...

A `100 Omega` resistance is connected in series with a `4 H` inductor. The voltage across the resistor is, `V_(R ) = (2.0 V) sin (10^(3) t)`.
Find the expression of circuit current

A

`(2 xx 10^(-2) A) sin (10^(3) t)`

B

`(2 xx 10^(-3) A) sin (10^(2) t)`

C

`(2 xx 10^(-3) A) sin (10^(3) t)`

D

`(2 xx 10^(-2) A) sin (10^(2) t)`

Text Solution

Verified by Experts

The correct Answer is:
A
`i = (V_(R ))/(R ) = ((2.0 v)sin (10^(3) t))/(100)`
`= (2.0 xx 10^(-2) A) sin (10^(3) t)`
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