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Standard XII
Physics
Introduction to Phasors and Problems on it
Question

Remove the inductor from the circuit in above figure, and set $$R=200\Omega$$, $$C=15.0\mu F$$, $$f_{d}=60.0Hz$$ and $$\xi _{m}=36.0V$$. What are (a) $$Z$$, (b) $$\phi$$ and (c) $$I$$? (d) Draw a phasor diagram.

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Solution
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(a) Now $$X_{L}=0$$, while $$R = 200 Ω$$ and $$X_{C}=1/2\pi f_{d}C=177\Omega$$. Therefore, the impedance is
$$Z=\sqrt{R^{2}+X_{C}^{2}}=\sqrt{\left ( 200\Omega \right )^{2}+\left ( 177\Omega \right )^{2}}=267\Omega$$.
(b) The phase angle is,
$$\phi=\tan ^{-1}\left ( \frac{X_{L}-C_{C}}{R} \right )=\tan ^{-1}\left ( \frac{0-177\Omega}{200\Omega} \right )=-41.5^{o}$$.
(c) The current amplitude is,
$$I=\frac{\varepsilon _{m}}{Z}=\frac{36.0V}{267\Omega}=0.135A$$.

(d) We first find the voltage amplitudes across the circuit elements:
$$V_{R}=IR=\left ( 0.135A \right )\left ( 200\Omega\right )\approx 27.0V$$,
$$V_{C}=IX_{C}=\left ( 0.135A \right )\left ( 177\Omega\right )\approx 23.9V$$.
The circuit is capacitive, so $$I$$ leads $$\varepsilon _{m}$$. The phasor diagram is drawn to scale next.

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