Answer :
To solve this problem, we need to find several electrical parameters for a circuit that consists of a resistor in series with a capacitor. We are given the following information:
- Power [tex]\( P = 100 \)[/tex] watts
- Power factor [tex]\( \text{pf} = 0.5 \)[/tex]
- Voltage [tex]\( V = 100 \)[/tex] volts
- Frequency [tex]\( f = 60 \)[/tex] Hz
Let’s proceed step-by-step:
### (a) Current Flowing (I)
The power [tex]\( P \)[/tex] in an AC circuit is given by the formula:
[tex]\[ P = V \cdot I \cdot \text{pf} \][/tex]
Rearranging to solve for the current [tex]\( I \)[/tex]:
[tex]\[ I = \frac{P}{V \cdot \text{pf}} \][/tex]
Substituting the given values:
[tex]\[ I = \frac{100}{100 \cdot 0.5} = 2.0 \, \text{A} \][/tex]
So, the current flowing is [tex]\( 2.0 \)[/tex] amperes.
### (b) Phase Angle (θ)
The power factor [tex]\( \text{pf} \)[/tex] is related to the phase angle [tex]\( \theta \)[/tex] by the cosine function:
[tex]\[ \text{pf} = \cos(\theta) \][/tex]
Therefore:
[tex]\[ \theta = \cos^{-1}(0.5) = \frac{\pi}{3} \approx 1.047 \, \text{radians} \][/tex]
So, the phase angle is approximately [tex]\( 1.047 \)[/tex] radians.
### (c) Resistance (R)
The power [tex]\( P \)[/tex] can also be expressed in terms of current and resistance:
[tex]\[ P = I^2 \cdot R \][/tex]
Rearranging to solve for the resistance [tex]\( R \)[/tex]:
[tex]\[ R = \frac{P}{I^2} \][/tex]
Substituting the given values:
[tex]\[ R = \frac{100}{2.0^2} = \frac{100}{4} = 25 \, \Omega \][/tex]
So, the resistance is [tex]\( 25 \)[/tex] ohms.
### (d) Impedance (Z)
The impedance [tex]\( Z \)[/tex] in an AC circuit can be calculated using Ohm's Law:
[tex]\[ Z = \frac{V}{I} \][/tex]
Substituting the given values:
[tex]\[ Z = \frac{100}{2.0} = 50 \, \Omega \][/tex]
So, the impedance is [tex]\( 50 \)[/tex] ohms.
### (e) Capacitance (C)
First, we need to find the reactance [tex]\( X \)[/tex] of the capacitor. The impedance [tex]\( Z \)[/tex] in a series RC circuit is related to resistance [tex]\( R \)[/tex] and reactance [tex]\( X \)[/tex] by:
[tex]\[ Z = \sqrt{R^2 + X^2} \][/tex]
Rearranging to solve for the reactance [tex]\( X \)[/tex]:
[tex]\[ X = \sqrt{Z^2 - R^2} \][/tex]
Substituting the values:
[tex]\[ X = \sqrt{50^2 - 25^2} = \sqrt{2500 - 625} = \sqrt{1875} \approx 43.30 \, \Omega \][/tex]
Now, the reactance [tex]\( X \)[/tex] of a capacitor is given by:
[tex]\[ X = \frac{1}{2\pi f C} \][/tex]
Rearranging to solve for the capacitance [tex]\( C \)[/tex]:
[tex]\[ C = \frac{1}{2\pi f X} \][/tex]
Substituting the values:
[tex]\[ C = \frac{1}{2 \pi \cdot 60 \cdot 43.30} \approx 6.13 \times 10^{-5} \, \text{F} \][/tex]
So, the capacitance is approximately [tex]\( 6.13 \times 10^{-5} \)[/tex] farads.
### Summary
1. Current flowing (I): [tex]\( 2.0 \)[/tex] amperes
2. Phase angle (θ): [tex]\( 1.047 \)[/tex] radians
3. Resistance (R): [tex]\( 25 \)[/tex] ohms
4. Impedance (Z): [tex]\( 50 \)[/tex] ohms
5. Capacitance (C): [tex]\( 6.13 \times 10^{-5} \)[/tex] farads
- Power [tex]\( P = 100 \)[/tex] watts
- Power factor [tex]\( \text{pf} = 0.5 \)[/tex]
- Voltage [tex]\( V = 100 \)[/tex] volts
- Frequency [tex]\( f = 60 \)[/tex] Hz
Let’s proceed step-by-step:
### (a) Current Flowing (I)
The power [tex]\( P \)[/tex] in an AC circuit is given by the formula:
[tex]\[ P = V \cdot I \cdot \text{pf} \][/tex]
Rearranging to solve for the current [tex]\( I \)[/tex]:
[tex]\[ I = \frac{P}{V \cdot \text{pf}} \][/tex]
Substituting the given values:
[tex]\[ I = \frac{100}{100 \cdot 0.5} = 2.0 \, \text{A} \][/tex]
So, the current flowing is [tex]\( 2.0 \)[/tex] amperes.
### (b) Phase Angle (θ)
The power factor [tex]\( \text{pf} \)[/tex] is related to the phase angle [tex]\( \theta \)[/tex] by the cosine function:
[tex]\[ \text{pf} = \cos(\theta) \][/tex]
Therefore:
[tex]\[ \theta = \cos^{-1}(0.5) = \frac{\pi}{3} \approx 1.047 \, \text{radians} \][/tex]
So, the phase angle is approximately [tex]\( 1.047 \)[/tex] radians.
### (c) Resistance (R)
The power [tex]\( P \)[/tex] can also be expressed in terms of current and resistance:
[tex]\[ P = I^2 \cdot R \][/tex]
Rearranging to solve for the resistance [tex]\( R \)[/tex]:
[tex]\[ R = \frac{P}{I^2} \][/tex]
Substituting the given values:
[tex]\[ R = \frac{100}{2.0^2} = \frac{100}{4} = 25 \, \Omega \][/tex]
So, the resistance is [tex]\( 25 \)[/tex] ohms.
### (d) Impedance (Z)
The impedance [tex]\( Z \)[/tex] in an AC circuit can be calculated using Ohm's Law:
[tex]\[ Z = \frac{V}{I} \][/tex]
Substituting the given values:
[tex]\[ Z = \frac{100}{2.0} = 50 \, \Omega \][/tex]
So, the impedance is [tex]\( 50 \)[/tex] ohms.
### (e) Capacitance (C)
First, we need to find the reactance [tex]\( X \)[/tex] of the capacitor. The impedance [tex]\( Z \)[/tex] in a series RC circuit is related to resistance [tex]\( R \)[/tex] and reactance [tex]\( X \)[/tex] by:
[tex]\[ Z = \sqrt{R^2 + X^2} \][/tex]
Rearranging to solve for the reactance [tex]\( X \)[/tex]:
[tex]\[ X = \sqrt{Z^2 - R^2} \][/tex]
Substituting the values:
[tex]\[ X = \sqrt{50^2 - 25^2} = \sqrt{2500 - 625} = \sqrt{1875} \approx 43.30 \, \Omega \][/tex]
Now, the reactance [tex]\( X \)[/tex] of a capacitor is given by:
[tex]\[ X = \frac{1}{2\pi f C} \][/tex]
Rearranging to solve for the capacitance [tex]\( C \)[/tex]:
[tex]\[ C = \frac{1}{2\pi f X} \][/tex]
Substituting the values:
[tex]\[ C = \frac{1}{2 \pi \cdot 60 \cdot 43.30} \approx 6.13 \times 10^{-5} \, \text{F} \][/tex]
So, the capacitance is approximately [tex]\( 6.13 \times 10^{-5} \)[/tex] farads.
### Summary
1. Current flowing (I): [tex]\( 2.0 \)[/tex] amperes
2. Phase angle (θ): [tex]\( 1.047 \)[/tex] radians
3. Resistance (R): [tex]\( 25 \)[/tex] ohms
4. Impedance (Z): [tex]\( 50 \)[/tex] ohms
5. Capacitance (C): [tex]\( 6.13 \times 10^{-5} \)[/tex] farads
