Physics · Alternating Current
JEE Main 2024 — 1 February, Shift 1 — Question 40
In series LCR circuit, the capacitance is changed from C to 4 C . To keep the resonance frequency unchanged, the new inductance should be :
- Option A:
reduced by
- Option B:
increased by 2 L
- Option C:Correct
reduced by
- Option D:
increased to 4L
Answer: C
Step-by-step solution
\begin{array}{*{35}{r}}{} & \frac{1}{\sqrt{\text{ }\!\!~\!\!\text{ }{{\text{L}}^{'}}\text{{C}'}}}=\frac{1}{\sqrt{\text{LC}}} \\{} & ~\therefore \text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }{{\text{L}}^{'}}\text{{C}'}=\text{LC} \\{} & \text{ }\!\!~\!\!\text{ }{{\text{L}}^{'}}\left( 4\text{C} \right)=\text{LC} \\{} & \text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }{{\text{L}}^{'}}=\frac{\text{L}}{4} \\\end{array}
Inductance must be decreased by
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2024
- Paper
- 1 February, Shift 1
- Subject
- Physics
- Chapter
- Alternating Current
- Topic
- Series LCR Circuit and Power Factor