>
Exams
>
Physics
>
Current electricity
>
in a series lcr circuit at resonance the voltage a
Question:
In a series LCR circuit at resonance, the voltage across the inductor is 100 V and across the capacitor is 100 V. The voltage across the resistor is:
Show Hint
At resonance, circuit impedance is purely resistive.
VITEEE - 2025
VITEEE
Updated On:
Jan 5, 2026
0 V
100 V
200 V
141 V
Hide Solution
Verified By Collegedunia
The Correct Option is
B
Solution and Explanation
At resonance, inductive and capacitive voltages are equal and opposite, so they cancel. The applied voltage appears entirely across the resistor.
Download Solution in PDF
Was this answer helpful?
0
0
Top Questions on Current electricity
Write the S.I. unit of: (i) Electric potential difference (ii) Resistance
TBSE Class X Board - 2026
Physics
Current electricity
View Solution
Why is one of the pins of a three-pin plug relatively thicker?
TBSE Class X Board - 2026
Physics
Current electricity
View Solution
State one advantage of using an MCB.
TBSE Class X Board - 2026
Physics
Current electricity
View Solution
What is the function of the earth wire?
TBSE Class X Board - 2026
Physics
Current electricity
View Solution
State Joule's laws of heating regarding electric current.
TBSE Class X Board - 2026
Physics
Current electricity
View Solution
View More Questions
Questions Asked in VITEEE exam
Find the value of \( x \) in the equation \( 2x + 3 = 7x - 8 \).
VITEEE - 2025
Algebra
View Solution
A convex lens has a focal length of 10 cm. What is the magnification produced when the object is placed 30 cm from the lens?
VITEEE - 2025
spherical lenses
View Solution
A ball is dropped from a height of 20 m. What is its velocity just before hitting the ground? (Take \( g = 9.8 \, \text{m/s}^2 \))
VITEEE - 2025
Motion in a straight line
View Solution
Solve for \( x \) in the equation \( \frac{2x - 3}{4} = 5 \).
VITEEE - 2025
Linear Equations
View Solution
Find the value of \( x \) in the equation \( 4(x - 2) = 3(x + 5) \).
VITEEE - 2025
Linear Equations
View Solution
View More Questions