In a series L – C – R circuit the frequency of a 10 V AC voltage source is adjusted in such a fashion that the reactance of the inductor measures $15\ \Omega$ and that of the capacitor $11\ \Omega$. If $R = 3\ \Omega$, the potential difference across the series combination of L and C will be

Question

Dr. Abubacker · Accepted Answer

(a) Given, $R = 3\Omega,\ {\ X}_{L} = 15\ \Omega,\ \ X_{C} = 11\ \Omega$
$V_{rms} = 10\ $volt
$	herefore$ Current through the circuit
$i = \frac{V_{rms}}{\sqrt{R^{2} + \left( X_{L} - X_{C} ight)^{2}}}$
$= \frac{10}{\sqrt{(3)^{2} + (15 - 11)^{2}}}$
$= \frac{10}{\sqrt{9 + 16}} = \frac{10}{5} = 2$A
Since, L, C and R are connected in series combination then potential difference across R is
${\ \ \ \ \ \ V}_{R} = i 	imes R = 2 	imes 3 = 6\ $V
Across L, ${\ \ \ \ \ \ V}_{L} = iX_{L} = 2 	imes 15 = 30\ $V
Across C, ${\ \ \ \ \ \ V}_{C} = iX_{C} = 2 	imes 11 = 22\ $V
So, potential difference across series combination of L and C,
$\ \ \ \ \ \ \ \ \ \ \  = V_{L} - V_{C}$
$\ \ \ \ \ \ \ \ \ \  = 30 - 22 = 8\ $V

Q. In a series L – C – R circuit the frequency of a 10 V AC voltage source is adjusted in such a fashion that the reactance of the inductor measures $15\ \Omega$ and that of the capacitor $11\ \Omega$. If $R = 3\ \Omega$, the potential difference across the series combination of L and C will be

Explanation

Reviewed by Dr. Abubacker

Want more practice?