Question

An alternating voltage given by \( e = 140 \sin (314.2 t) \) is connected across a pure resistor of 50 \(\Omega\). Calculate:
(i) the frequency of the source
(ii) the r.m.s current through the resistor

Hint:

The r.m.s. value of current in a resistor is the ratio of the peak voltage to the resistance, and the frequency is found from the angular frequency.

Answer 1

Step 1: Find the frequency of the source.
The given equation for voltage is \( e = 140 \sin (314.2 t) \), which is of the form: \[ e = E_0 \sin (\omega t) \] where \( E_0 = 140 \) is the amplitude and \( \omega = 314.2 \) is the angular frequency. The angular frequency is related to the frequency \( f \) by: \[ \omega = 2 \pi f \] Substitute \( \omega = 314.2 \): \[ 314.2 = 2 \pi f \] Solving for \( f \): \[ f = \frac{314.2}{2 \pi} \approx 50 \, \text{Hz} \]
Step 2: Find the r.m.s current through the resistor.
The r.m.s. current \( I_{\text{rms}} \) in a resistor is given by: \[ I_{\text{rms}} = \frac{E_0}{R} \] where \( R = 50 \, \Omega \) is the resistance and \( E_0 = 140 \, \text{V} \) is the amplitude of the voltage. Substituting the values: \[ I_{\text{rms}} = \frac{140}{50} = 2.8 \, \text{A} \]