Electrical Engineering - RC Circuits - Discussion
Discussion Forum : RC Circuits - General Questions (Q.No. 26)
26.
A capacitor with 150 
of capacitive reactance is across an ac source. The impedance, expressed in polar form, is
of capacitive reactance is across an ac source. The impedance, expressed in polar form, isZ = 150
Z = 150
90°
90°Z = 150 –90°
–90°Z = 150180°
180°Discussion:
5 comments Page 1 of 1.
HAris said:
6 years ago
Z = R-jxc as R which is not given, it's negligible
Z = -j150.
-For polar;
r = √ (x^2+y^2),
r = √ (+(-150)^2),
r = 150,
∠= tangent inverse(y/x),
=tan.inv(150),
= 90 = -90 and sign - because in capacitive impedance lags by 90.
Z = -j150.
-For polar;
r = √ (x^2+y^2),
r = √ (+(-150)^2),
r = 150,
∠= tangent inverse(y/x),
=tan.inv(150),
= 90 = -90 and sign - because in capacitive impedance lags by 90.
Arslan zahoor said:
1 decade ago
Current is leading to voltage 90 degree in capacitor and current is lagging to voltage 90 degree in inductor.
Amit said:
1 decade ago
Why it is -90 it should be +90?
Deepesh soni said:
1 decade ago
Current lead in capacitor by 90 so impedance will become -90.
Satya prakash said:
1 decade ago
Its very simple
1/jwc
Since j is in denominator it will lag by angle=-90degree
1/jwc
Since j is in denominator it will lag by angle=-90degree
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