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 Ω and that of the capacitor 11 Ω . If R=3 Ω the potential difference across the series combination of L, and C will be:

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 Ω and that of the capacitor 11 Ω . If R=3 Ω 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

Tardigrade Admin · Accepted Answer

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 _{r m s}}{R ^{2} + ( X _{L} - X _{c} ) ^{2}}

Given,

R = 3 Ω, X_{L} = 15 Ω, X_{C} = 11 Ω,

V_{r m s} = 10 V

∴

i = \frac{10}{( 3 ) ^{2} + ( 15 - 11 ) ^{2}}

Or

i = \frac{10}{9 + 16}

Or

i = \frac{10}{5} = 2 A

Since L, C and R are connected in series combination, then potential difference across R is

V_{R} = i R = 2 \times 3 = 6 V

Across

L, V_{L} = i X_{L} = 2 \times 15 = 30 V

Across

C, V_{C} = i X_{C} = 2 \times 11 = 22 V

So, potential difference across series combination of L and C

= V_{L} - V_{C}

= 30 - 22

= 8 V

Note: In

L CR

circuit whenever voltage across various elements is asked, find rms values unless stated in the question for the peak or instantaneous value.

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

8 V

10 V

22 V

52 V

Q. 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 $15 Ω$ and that of the capacitor $11 Ω$ . If $R = 3 Ω$ the potential difference across the series combination of $L$ , and $C$ will be: