# Alternate current problem

Discussion in 'Homework Help' started by ghoasted, Feb 11, 2010.

1. ### ghoasted Thread Starter New Member

Feb 11, 2010
1
0
See picture to see the circuit. Both resistors are worth R.
V(t) is an emf with angular frequency $\omega$ and amplitude $V_0$.
1)Find the current that passes through the emf.
2)Let $V_{AB}$ be worth $V_B-V_A$. Demonstrate that $|V_{AB}|^2=V_0^2$ for all $\omega$.
3)Find the angle difference of phase between the current passing through the emf and the current reaching a capacitor.

My attempt:
$V(t)=V_0 \cos (\omega t)$. Also, $V(t)=Zi(t)$ where Z is the impedance of the circuit.
As both branches of the circuit are in parallel, the admittance of both branches sum up. $G=\frac{1}{Z}$.
So $G=G_1+G_2 \Rightarrow Z=\frac{1}{G}=\frac{1}{G_1+G_2}=\frac{1}{\left ( \frac{1}{Z_1}+\frac{1}{Z_2} \right )}$ where $Z_1$ and $Z_2$ are the impedances of each branch, respectively.
But $Z_1=Z_2$. Thus $Z=\frac{Z_1}{2}$.
$Z_1=R-\frac{1}{i\omega C}$, thus $Z=\frac{R}{2}+\frac{i}{2\omega C}$.
Finally, $i(t)=\frac{2V_0 \omega C \cos (\omega t)}{R\omega C+i}$, where the i in the denominator is the complex number.
Am I right for question 1)? Did I make an error?
For 2), I'm not really sure about how to proceed.
I believe that $V_B-V_A=\frac{i(t)R}{4}-\frac{q(t)}{C}$. But I'm unsure and even if it were right, I don't know how to continue. I know that $i=\frac{dq(t)}{dt}$, but does this help?

Any help is greatly appreciated! Thanks in advance.

File size:
57.2 KB
Views:
22
2. ### mik3 Senior Member

Feb 4, 2008
4,846
67
Find the impedance of all the elements in the circuit. Then find Va and Vb with the voltage divider rule or whatever you like. Then do Va-Vb to find the answer.