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Sagot :
Alright, let’s solve the problem step-by-step.
### Given:
- The voltage equation: [tex]\( e = 140 \sin 3142t \)[/tex]
- A pure resistor [tex]\( R = 50 \, \Omega \)[/tex]
### Part (i): Finding the frequency of the source
We start with the given equation for the voltage:
[tex]\[ e = 140 \sin 3142t \][/tex]
The general form of an alternating voltage is given by:
[tex]\[ e = E_{\text{max}} \sin(2 \pi f t) \][/tex]
In this form, [tex]\( E_{\text{max}} \)[/tex] is the maximum voltage, and [tex]\( 2 \pi f \)[/tex] is the angular frequency.
We compare the given equation [tex]\( e = 140 \sin 3142t \)[/tex] to the general form [tex]\( e = E_{\text{max}} \sin(2 \pi f t) \)[/tex]:
- [tex]\( 2 \pi f \)[/tex] corresponds to [tex]\( 3142 \)[/tex]
From this, we can solve for [tex]\( f \)[/tex] (the frequency):
[tex]\[ 2 \pi f = 3142 \][/tex]
[tex]\[ f = \frac{3142}{2 \pi} \][/tex]
Using the numerical result we have:
[tex]\[ f \approx 500.0648311947352 \, \text{Hz} \][/tex]
So, the frequency of the source is approximately [tex]\( 500.065 \, \text{Hz} \)[/tex].
### Part (ii): Finding the RMS current through the resistor
First, we need to find the RMS (Root Mean Square) value of the voltage. For a sinusoidal voltage, the RMS value is given by:
[tex]\[ E_{\text{rms}} = \frac{E_{\text{max}}}{\sqrt{2}} \][/tex]
Given:
- [tex]\( E_{\text{max}} = 140 \, \text{V} \)[/tex]
We calculate [tex]\( E_{\text{rms}} \)[/tex]:
[tex]\[ E_{\text{rms}} = \frac{140}{\sqrt{2}} \][/tex]
Using the numerical result we have:
[tex]\[ E_{\text{rms}} \approx 98.99494936611664 \, \text{V} \][/tex]
Next, we need to find the RMS current through the resistor. Ohm's law states:
[tex]\[ I_{\text{rms}} = \frac{E_{\text{rms}}}{R} \][/tex]
Given:
- [tex]\( E_{\text{rms}} \approx 98.99494936611664 \, \text{V} \)[/tex]
- [tex]\( R = 50 \, \Omega \)[/tex]
We calculate [tex]\( I_{\text{rms}} \)[/tex]:
[tex]\[ I_{\text{rms}} = \frac{98.99494936611664}{50} \][/tex]
Using the numerical result we have:
[tex]\[ I_{\text{rms}} \approx 1.979898987322333 \, \text{A} \][/tex]
So, the RMS current through the resistor is approximately [tex]\( 1.98 \, \text{A} \)[/tex].
### Summary
(i) The frequency of the source is approximately [tex]\( 500.065 \, \text{Hz} \)[/tex].
(ii) The RMS current through the resistor is approximately [tex]\( 1.98 \, \text{A} \)[/tex].
### Given:
- The voltage equation: [tex]\( e = 140 \sin 3142t \)[/tex]
- A pure resistor [tex]\( R = 50 \, \Omega \)[/tex]
### Part (i): Finding the frequency of the source
We start with the given equation for the voltage:
[tex]\[ e = 140 \sin 3142t \][/tex]
The general form of an alternating voltage is given by:
[tex]\[ e = E_{\text{max}} \sin(2 \pi f t) \][/tex]
In this form, [tex]\( E_{\text{max}} \)[/tex] is the maximum voltage, and [tex]\( 2 \pi f \)[/tex] is the angular frequency.
We compare the given equation [tex]\( e = 140 \sin 3142t \)[/tex] to the general form [tex]\( e = E_{\text{max}} \sin(2 \pi f t) \)[/tex]:
- [tex]\( 2 \pi f \)[/tex] corresponds to [tex]\( 3142 \)[/tex]
From this, we can solve for [tex]\( f \)[/tex] (the frequency):
[tex]\[ 2 \pi f = 3142 \][/tex]
[tex]\[ f = \frac{3142}{2 \pi} \][/tex]
Using the numerical result we have:
[tex]\[ f \approx 500.0648311947352 \, \text{Hz} \][/tex]
So, the frequency of the source is approximately [tex]\( 500.065 \, \text{Hz} \)[/tex].
### Part (ii): Finding the RMS current through the resistor
First, we need to find the RMS (Root Mean Square) value of the voltage. For a sinusoidal voltage, the RMS value is given by:
[tex]\[ E_{\text{rms}} = \frac{E_{\text{max}}}{\sqrt{2}} \][/tex]
Given:
- [tex]\( E_{\text{max}} = 140 \, \text{V} \)[/tex]
We calculate [tex]\( E_{\text{rms}} \)[/tex]:
[tex]\[ E_{\text{rms}} = \frac{140}{\sqrt{2}} \][/tex]
Using the numerical result we have:
[tex]\[ E_{\text{rms}} \approx 98.99494936611664 \, \text{V} \][/tex]
Next, we need to find the RMS current through the resistor. Ohm's law states:
[tex]\[ I_{\text{rms}} = \frac{E_{\text{rms}}}{R} \][/tex]
Given:
- [tex]\( E_{\text{rms}} \approx 98.99494936611664 \, \text{V} \)[/tex]
- [tex]\( R = 50 \, \Omega \)[/tex]
We calculate [tex]\( I_{\text{rms}} \)[/tex]:
[tex]\[ I_{\text{rms}} = \frac{98.99494936611664}{50} \][/tex]
Using the numerical result we have:
[tex]\[ I_{\text{rms}} \approx 1.979898987322333 \, \text{A} \][/tex]
So, the RMS current through the resistor is approximately [tex]\( 1.98 \, \text{A} \)[/tex].
### Summary
(i) The frequency of the source is approximately [tex]\( 500.065 \, \text{Hz} \)[/tex].
(ii) The RMS current through the resistor is approximately [tex]\( 1.98 \, \text{A} \)[/tex].
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