$A$ radio can tune over the frequency range of a portion of $MW$ broadcast band: ($800 \;kHz$ to $1200 \;kHz$). If its $LC$ circuit has an effective inductance of $200 \;\mu H$,what must be the range of its variable capacitor?
- A$88 \;pF$ to $198 \;pF$
- B$198 \;pF$ to $88 \;pF$
- C$100 \;pF$ to $200 \;pF$
- D$50 \;pF$ to $150 \;pF$
Explore More
Similar Questions
An inductor of $\left(\frac{100}{\pi}\right) mH$,a capacitor of capacitance $\left(\frac{10^{-3}}{2 \pi}\right) F$,and a resistor of $10 \Omega$ are connected in series with an $AC$ voltage source of $110 \text{ V}, 50 \text{ Hz}$. The tangent of the phase angle $\phi$ between the voltage and the current is:
$A$ resistor and an inductor are connected in series to an $AC$ source of voltage $V = 150 \sin (100 \pi t + \pi) \text{ V}$. If the current in the circuit is $I = 5 \sin (100 \pi t + \frac{2 \pi}{3}) \text{ A}$,then the average power dissipated and the resistance of the resistor are respectively:
$A$ resistance of $40\, \Omega$ and an inductance of $95.5\, \text{mH}$ are connected in series in a $50\, \text{Hz}$ $AC$ circuit. The impedance of this combination is very nearly.....$\Omega$.
Easy
View SolutionIn a series $RLC$ circuit,the potential differences across $R, L$ and $C$ are $30 \,V, 60 \,V$ and $100 \,V$ respectively,as shown in the figure. The $e.m.f.$ of the source (in volts) is:
Easy
View Solution$A$ coil of $200 \, \Omega$ resistance and $1.0 \, H$ inductance is connected to an $ac$ source of frequency $200/2\pi \, Hz.$ The phase angle between potential and current will be.....$^o$
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo