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An alternating voltage V = 100 sin omega...

An alternating voltage `V = 100 sin omega t` is applied across an `LCR` circuit as shown. At the instant when voltage drop across source is `50 sqrt(3)` volts then at that instant
`R = 30 Omega" "X_(L ) = 60 Omega" "X_(C ) = 20 Omega`

A

voltage drop across inductor is `(120 cos 7^(@))` volts

B

voltage drop across capacitor is `(40 cos 173^(@))` volts

C

voltage drop across resistor is `(60 cos 7^(@))` volts

D

All the above

Text Solution

Verified by Experts

The correct Answer is:
D

`i_(0) = (V_(0))/(sqrt(R^(2) + (X_(L) - X_(C ))^(2)) )= (100)/(50) = 2 A`
Peak value of voltages
`|vec(V_(R ))| - i_(0) R = 60 V`
`|vec(V_(L))| = i_(0) X_(L) = 120 V`
`|vec(V_(C))| = i_(0) X_(C ) = 40 V`
Phasor diagram at the instant given
Instantaneous values are components on + y-axix.
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