Question

In a series LCR 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:

• A. 8 V
• B. 10 V
• C. 22 V
• D. 52 V

Answer 1

Answer: (A)

The potential difference across the series combination of L and C is the difference in potential differences across individuals. Current through the circuit,
$ i=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{c}})}^{2}}}} $
Given, $ R=3\,\Omega ,\,{{X}_{L}}=15\,\Omega ,\,{{X}_{C}}=11\,\Omega , $
$ {{V}_{rms}}=10\,V $
$ \therefore $ $ i=\frac{10}{\sqrt{{{(3)}^{2}}+{{(15-11)}^{2}}}} $
Or $ i=\frac{10}{\sqrt{9+16}} $
Or $ i=\frac{10}{5}=2A $
Since L, C and R are connected in series combination, then potential difference across R is
$ {{V}_{R}}=iR=2\times 3=6V $
Across $ L,{{V}_{L}}=i{{X}_{L}}=2\times 15=30V $
Across $ C,{{V}_{C}}=i{{X}_{C}}=2\times 11=22V $
So, potential difference across series combination of L and C
$={{V}_{L}}-{{V}_{C}} $
$=30-22 $
$=8\,V $
Note: In $ LCR $ circuit whenever voltage across various elements is asked, find rms values unless stated in the question for the peak or instantaneous value.