Recent content by jojo13

Ahh I get it. So no matter what, a 50 Hz can be fed to a 60 Hz

Will a transformer with 60 Hz being fed to a 50 Hz supply work?

Thanks

Ah, would it be, Vc = I * Zc = 10 * (1/j) = 10/j = -10j

Yeah, I was thinking it would then go up

If a motor slowed to a halt, will the current go up or down

Homework Statement 10 v, 100 hz goes into a circuit of a 1o resistor, a 1o inductive reactance and a 1o capacitive reactance that are in series. What is the current. What is the V across the cap. Homework Equations and the attempt at a solution[/B] So I know I = V/Z and Z = sqrt( R^2 +...

-VC + 7V + V1 = 0. VC = 3 V

Hmm, I must have done a sign error or something, this is the work I did for it I first found IL which was -4 A. Then I did KVL on the left side which I found to be -VC + 7V - V1 = 0. I subsituted V1 to be (-4 A)*(1 ohm) and solved for the rest. Ended up with 11V.

ok I think I got the final answers. Mind checking if I did them right. I was a bit confused on the open circuit at VC, so I wanted to make sure it's correct. At DCSS, replace inductor with short circuit At DCSS, replace capacitor with open circuit For t = 0- DCSS, I got IL = 0A and VC = 2V...

Ahhh I see. So just to confirm... At both t = 0 and t = ∞, the left side is 7V and the right side is 4V. That makes a lot more sense than what I was thinking.

Ok thanks, so for the t = 0- DCSS, the 2+5u(t) becomes 2v and the 4u(t) becomes a short. Now for the t = ∞, the 2+5u(t) becomes 7V and the 4u(t) becomes 4V? I was thinking that the 2+5u(t) becomes 2V and the 4u(t) becomes 4V

Homework Statement Two independent sources are given. They are active at different times. Find the conditions at t = 0- DCSS (aka t < 0 DCSS) and t = ∞ DCSS for IL and Vc Homework Equations At DCSS, replace inductor with short circuit At DCSS, replace capacitor with open circuit The...

I see. That's and interesting way of looking at the problem. Thanks for that. Also, if I'm not mistaken both give the same answer (∏/2) right?

Homework Statement Find the area of the region that consists of all points that lie within the circle r = 1 but outside the polar equation r = cos(2θ) Homework Equations A = ∫ 1/2 (r2^2 - r1^2) dθ, where r2 is outer curve and r1 is inner curve. The Attempt at a Solution Here is...