In above figure, set $$R=200\Omega$$, $$C=70.0\mu F$$, $$L=230 mH$$, $$f_{d}=60.0Hz$$ and $$\xi _{d}=36.0V$$. What are (a) $$Z$$, (b) $$\phi$$ and (c) $$I$$? (d) Draw a phasor diagram.

(a) The capacitive reactance is,

$$X_{C}=\frac{1}{\omega _{d}C}=\frac{1}{2\pi f_{d}C}=\frac{1}{2\pi \left ( 60.0Hz \right )\left ( 70.0*10^{-6}F \right )}=37.9\Omega$$.

The inductive reactance $$86.7 Ω$$ is unchanged. The new impedance is

$$Z=\sqrt{R^{2}+\left ( X_{L}-X_{C} \right )^{2}}=\sqrt{\left ( 200\Omega\right )^{2}+\left ( 37.9\Omega -86.7\Omega\right )^{2}}=206\Omega$$.

(b) The phase angle is,

$$\phi=\tan ^{-1}\left ( \frac{X_{L}-X_{C}}{R} \right )=\tan ^{-1}\left ( \frac{86.7\Omega -37.9\Omega }{200\Omega} \right )=13.7^{0}$$.

(c) The current amplitude is,

$$I=\frac{\varepsilon _{m}}{Z}=\frac{36.0V}{206\Omega}=0.175A$$.

(d) We first find the voltage amplitudes across the circuit elements:
$$V_{R}=IR=\left ( 0.175A \right )\left ( 200\Omega \right )=35.0V$$.
$$V_{L}=IX_{L}=\left ( 0.175A \right )\left ( 86.7\Omega \right )=15.2V$$.
$$V_{C}=IX_{C}=\left ( 0.175A \right )\left ( 37.9\Omega \right )=6.62V$$.
Note that $$X_{L}>X_{C}$$, so that $$\varepsilon _{m}$$ leads $$I$$. The phasor diagram is drawn to scale above.