Recent content by stepfanie

thank you nascent :D you help alot..!

isn't v^2 / 2R ?

Is it like how I did in the previous question finding the real denominator and imn thingy ?

So how can we determine the total active power? I have tried but only come out with that. :(

this is what i got as well.

thank you. but if i want to find the total active power ? what will it be.. and can u solve d)? I am not sure I am doing correctly.

for c) we not need to calculate the total P ? just P in each resistance ? :rolleyes: and for b) we just use the magnitude of Vab and how about the XL ?

for d) Q=|V^2| / XL v= (jXL / (R + jXL)) *v -> voltage divider rule. v= (jVXL)/ (R+ jXL) v^2 = [ (jVXL)/ (R+ jXL) ] ^2 v = (j^2 V^2 XL^2) / ((R + jXL)^2) v= (-V^2 XL^2) / ((R + jXL)^2) so sub V into Q=|V^2| / XL Q= [(-V^2 XL^2) / ((R + jXL)^2)] / XL Q= (-V^2 XL^2) / (XL*((R +...

c) active power dissipated in each resisance formula p=(I^2)*R [( V∠0 (3R+Xl∠90 )) / ( 3R+2RXl∠90 )]^2 *R d) reactive power dissipated in XL Q=I^2 * jXL [( V∠0 (3R+Xl∠90 )) / ( 3R+2RXl∠90 )]^2 * Xl∠90 so for Power Factor P/Q = [(( V∠0 (3R+Xl∠90 )) / ( 3R+2RXl∠90 ))^2 *R] / [(( V∠0...

for f) calculate the power factor of the circuit. power factor = true power / apparent power which is (I^2 *R) / (I^2 *Z) so i can use the equation in c) which is the active power dissipated in each resistance divide by d) which is the reactive power dissipated in XL by using power factor...

so the magnitude = V sqrt of ((R^2-XL^2)/(2R^2+2XL^2))^2 + (-RXL/(R^2+XL^2)) ^2 ) that's it right? :D

is it the denominator (R^2 + XL^2) ?

sqrt of ( (R^2-XL^2)^2+ (-2jRXL)^2) but the angle.

magnitude is (-2jRXL) / (R^2+XL^2) Vertical = imaginary angle is (v/2) * [ (R^2 - XL^2)/(R^2+XL^2) Horizontal = real

just wonder is 1 or -1 ? if -1 then 63.43°