Physics Answer the following in Brief

An alternating emf, E = 200sin(314.2t) volt is applied  between the terminals of an electric bulb whose filament has a resistance of 100Ω.
We have to calculate the value of

1. A) RMS current
2. B) Frequency of AC signal
3. C) Period of AC signal

A) Here, alternating emf, E = 200sin(314.2t) = 200sin(100πt) [∵ 3.14 = π ]
Comparing to E = E₀sin(ωt)
We get, E₀ = 200 volt , ω = 100π
We know from Ohm's law, V = iR
\(i_0=\frac{E_0}{R}=\frac{200}{100}=2 A\)
[ here resistance of filament is R = 100Ω. ]
Now RMS current is given by,
\(I _{ rms }=\frac{ I _0}{\sqrt{2}}\)
\(=\frac{2}{\sqrt{2}}=\sqrt{2} A\)
Therefore the RMS current is √2 A.
B) We get, ω = 100π
We know, ω = 2πη , here η is the frequency of AC signal.
∴ 100π = 2πη
⇒ η = 50 Hz
Therefore the frequency of AC signal is 50 Hz.
C) We know, Period is the time taken to complete one cycle.
period, \(T=\frac{1}{\eta}=\frac{1}{50}=0.02 s\)
Therefore the period of AC signal is 0.02 sec.

The impedance of series LCR circuit is
\(Z=\sqrt{R^2+\left(X_L-X_c\right)^2}\)
where \(X_L\) and \(X_c\) are the inductive and capacitive reactances respectively.
At resonance \(X_L=X_c\)
So
(i) the impedance at resonance
\(Z_{r e s}=R=12 \Omega\)
(ii) the current at resonance
\(I_{\text {res }}=\frac{V}{Z_{\text {res }}}=\frac{V}{R}=\frac{120 V }{12 \Omega}=10 A\)
(iii) At resonance \(X_L=X_c\)
i.e.,
\(\omega L=\frac{1}{\omega C}\)
i.e.,
\(\omega^2=\frac{1}{L C}\)
i.e.,
\(\omega=\frac{1}{\sqrt{L C}}\)
So the frequency is
\(f=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{1000 \times 10^{-3} \times 36 \times 10^{-3}}} Hz\)
\(=0.8388 Hz\)
If 36 micro farad capacitor is used
f=26.5 Hz