Q.

A pure resistive circuit element X when connected to an AC supply of peak voltage 200V gives a peak current of 5A which is in phase with the voltage. A second circuit element Y, when connected to the same AC supply also gives the same value of peak current but the current lags behind by 900. If the series combination of X and Y is connected to the same supply, the rms value of current is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

102A

b

52A

c

52A

d

5A

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

for the first case  2005=40Ω=RIf current lags by 90o  it is an inductive circuit:  2005=40Ω=XLFor series impedance  = Z=XL2+R2=402irms=2002.402=52A
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
whats app icon
A pure resistive circuit element X when connected to an AC supply of peak voltage 200V gives a peak current of 5A which is in phase with the voltage. A second circuit element Y, when connected to the same AC supply also gives the same value of peak current but the current lags behind by 900. If the series combination of X and Y is connected to the same supply, the rms value of current is