Question

An alternating e.m.f. e = 200 sin 314.2t volt is applied between the terminals of an electric bulb whose filament has a resistance of 10052. Calculate the following:

(a) RMS current

(b) Frequency of AC signal

(c) Period of AC signal​

Answer 1

Answer:

(a)RMS current-0.013A

(b)Frequency of AC signal-50.03Hz

(c)Period of AC signal​-0.02s.

Explanation:

The emf(e) and resistance (R) given in the question is,

$e=200sin314.2tV$      (1)

$R=10052\Omega$               (2)

(a) The rms current(Irms) is given as,

$I_r_m_s=\frac{I_o}{\sqrt{2}}$                 (3)

I₀=peak current

So, first, let's find the peak current,

$I_o=\frac{e_o}{R}=\frac{200}{10052}=0.019A$       (e₀=200 from equation (1))    

By substituting the value of I₀ in equation (3) we get;

$I_r_m_s=\frac{0.019}{\sqrt{2}}=0.013A$

(b) From equation (1) we have angular frequency(ω)=314.2

And we know,

$\omega=2\pi f$         (4)

Where f is the frequency,

So,

$314.2=2\times 3.14\times f$

$f=\frac{314.2}{2\times 3.14}=\frac{100.06}{2}=50.03Hz$

(c) The period (T) of AC signal is given as,

$T=\frac{1}{f}$                 (6)

So,

$T=\frac{1}{50.03}=0.0199\approx0.02s$

Hence, the rms current is 0.013A, the frequency of the AC signal is 50.03Hz and the period of the signal is 0.02s.