Q. In the circuit shown in the figure, the ac source gives a voltage $V = 20\cos{(2000t)}$. Neglecting source resistance, the voltmeter and ammeter reading will be
A) $0V,\ 0.47A$
B) $1.68V,\ 0.47A$
C) $0V,\ 1.4\ A$
D) $5.6V,\ 1.4\ A$
Explanation
(d) $Z = \sqrt{(R)^{2} + \left( X_{L} - X_{C} \right)^{2};}$
$R = 10\Omega,\ X_{L} = \omega L = 2000 \times 5 \times 10^{- 3} = 10\Omega$
$X_{C} = \frac{1}{\omega C} = \frac{1}{2000 \times 50 \times 10^{- 6}} = 10\Omega,\ i.e.,\ Z = 10\Omega$
Maximum current $i_{0} = \frac{V_{0}}{Z} = \frac{20}{10} = 2A$
Hence $i_{rms} = \frac{2}{\sqrt{2}} = 1.4\ A$ and $V_{rms} = 4 \times 1.41 = 5.64\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.