# Current and Parallel Impedance

1. Oct 31, 2014

### Zondrina

1. The problem statement, all variables and given/known data

Find $i(t)$ in the following circuit:

2. Relevant equations

$Z = \frac{V}{I} \Rightarrow I = \frac{V}{Z}$

3. The attempt at a solution

I've solved this, but I'm wondering why my answer is different than the book's answer. The book lists the answer as $i(t) = 3.88 cos(377t - 39.2^o) \space A$.

Writing the given voltage as a phasor: $v(t) = 120 \angle 60^o \space V$.

Calculating the impedances of each component:

$Z_R = 20 \Omega$
$Z_L = j \omega L = (377)(40 \times 10^{-3})j = (15.08j) \Omega$
$Z_C = - \frac{j}{\omega C} = - \frac{j}{(377)(50 \times 10^{-6})} = -(53.05j) \Omega$

Calculating the required equivalent impedances:

$Z_{eq_1} = Z_R + Z_L = (20 + 15.08j) \Omega$

$Z_{eq_2} = (\frac{1}{Z_C} + \frac{1}{Z_{eq_1}})^{-1} = (- \frac{1}{53.05 j} + \frac{1}{20 + 15.08j})^{-1} = (30.56 + 4.97j) = 30.96 \angle 9.24^o \space \Omega$

Finding the current phasor:

$I = \frac{V}{Z} = \frac{120 \angle 60^o}{30.96 \angle 9.24^o} = 3.88 \angle 50.8^o \space A$

$i(t) = 3.88 cos(377t + 50.8^o) \space A$

My answer differs from the books answer by exactly $90^o$ (50.8 + 39.2 = 90). Why is this? Have I done something wrong?

2. Oct 31, 2014

### Staff: Mentor

3. Oct 31, 2014

### Zondrina

That's good to know. Although I'm a little perplexed as to why my answer differs by an angle of $\frac{\pi}{2}$.

4. Oct 31, 2014

### Staff: Mentor

Text books have been known to have incorrect solutions on occasion.