Recent content by Gabry89

done! I've found out how to manage that problem: (KCL) IE| = IR2| + IR4| + IR5| =10 A (KCL) IE|| = IR1| - IR3|| = 1 A PJ = J2*Req + (I3|*R3 - I1|*R1) * J = 1392 W PE = E*IE| - IE||*E

Sorry i wrote it wrong in the post, i wrote I3 instead of I4 and viceversa The correct currents are: I1 = I1| + I1|| = 2 A I2 = I2| + I2|| = 14 A I3 = I3| + I3|| = 7 A I4 = I4| + I4|| = 9 A yes, sorry, here's what i mean Req = (R1 || R2) + (R3 || R4) = 5,3125 \Omega PJ = J2 * Req = 1360 W

too bad i don't find the correct answer, i tried it again with my colleagues but I've found PJ = 1360 W rather than PJ = 1392 W which is the correct answer.

Sure, it's the current divider, i thought about that but i was confused by the short circuit in the middle, well thanks now it's clear. Well, when a resistor is in parallel with a short circuit the voltage across the resistor is 0 then the equivalent resistance of that combination is 0.

then how would you get the Req? and why if i consider them in series for calculating Req the result is correct? Talking about PJand PE i tried doing like this: PJ = J*Vab = 819.2 W PE= E*IE = E2 / Req = 320 which are not the correct results. The correct results are: PJ= 1392 W, PE = 288 W

Then it should be right. Now i have to find the power of the current source and the same for the voltage source. I've found that: I1 = I1| + I1|| = 2 A I2 = I2| + I2|| = 14 A I3 = I3| + I3|| = 9 A I4 = I4| + I4|| = 7 A Pe+PJ= I12R1 + I22R2 + I32R3 + I42R4 + E2/R5 But how i get...

i've studied that when I'm trying to get Req i don't have to consider the presence of the sources as they would have been switched off. According to this consideration here's why they're in series. Thank you very much, here's a better place. is that correct now? well, i know that these...

actually they're not draw in the same way and yes they're a voltage and a current source. Then i don't understand how to get the second component cause I've tried again the exercise and I've found the correct result thinkin' about R1 and R2 in parallel and the same for R3 and R4. I've...

Homework Statement I have to find I3, I've already easily found the first component I3| but i don't understand how i can get I3||. I thought about a Wheatstone bridge but doesn't work looking at the data. The result is I3=-7 A

Solved, it was not so difficult, anyway the formulas in the first post are wrong. You just have to be careful using the voltage divider and the KCL, KVL. The thread can be closed.

It is kinda difficult 'cause i must use the superposition. The results should be i1= 0,5 A and i2= 0,65 A

I tried it many times but i don't understand how to start and which are the first steps. I started like this: Solve for V1: Vr1 = V1 * (R1/R1+(R2||(R3+R4))) = 5 V I'1 = Vr1/R1 = 1 A I'2 = I'1 * [(R1+R2)||(R3+R4)]/(R3+R4) = 0,6 A. is that right?

i'm arrived at this integral: 1/12 * \int_{0}^{\pi/2}\ (cos\varphi)/[cos\varphi+sin\varphi]^3 d\varphi i've seen that it gives me 1/24 using a calculator so I've found the correct result but i just don't know how to get to 1/2 solving the integral.

i did \intfrom pi/2 to 2pi \int from 0 to 1 \int 0 to 1-\rhocos\varphi-\rhosin\varphi of [(\rho^2)cos\varphi] dy d\rho d\varphi which gives me -1/3 + 1/8 -(3pi)/16 so a part of the right result ( we can see -(3pi)/16)

Ok, even if the first one looks kinda complicated 'cause i get (talkin' about \rho) \int from 0 to 1/(sin\varphi+cos\varphi) of \rho^2 and besides the same for \rho^3, then is there an easier way to solve these integrals than e.g. calculating the cube? the second one gives me a part of the...