Question

The value of C is ___.

Answer 1

Correct Answer: 100

Step 1: Power in terms of voltage and resistance
The formula for power is given by:
\( P = \frac{V^2}{2} \)
We are given that the power \( P = 500 \) W, so:
\( 500 = \frac{100^2}{R} \)
Solving for \( R \), we get:
\( R = 20 \, \Omega \)

Step 2: Current calculation
Using Ohm's law across the resistor, we know:
\( P = I_{\text{rms}} \times V_{\text{rms}} \)
Given that \( P = 500 \) W and \( V_{\text{rms}} = 100 \) V, we can calculate the current \( I_{\text{rms}} \):
\[ 500 = I_{\text{rms}} \times 100 \] Solving for \( I_{\text{rms}} \), we get:
\( I_{\text{rms}} = 5 \, \text{A} \)

Step 3: Voltage and real power comparison
The RMS voltage across the resistance is given as \( V_{\text{rms}} = 200 \) V.
The real voltage (effective voltage) is 100 V, so the ratio is:
\( \frac{V_{\text{rms}}}{\text{real}} = 100 \, \text{V} \)

Step 4: Calculating the phase angle \( \varphi \)
The power factor \( \cos \varphi \) is the ratio of the real voltage to the RMS voltage:
\( \cos \varphi = \frac{100}{200} = \frac{1}{2} \)
Thus, the phase angle \( \varphi \) is:
\( \varphi = 60^\circ \)

Step 5: Calculate the reactance \( X_C \) and capacitance \( C \)
The relationship between reactance \( X_C \), resistance \( R \), and the phase angle \( \varphi \) is given by:
\( \tan \varphi = \frac{X_C}{R} = \frac{1}{\omega R C} \)
Using the value of \( \tan 60^\circ = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{1}{100\pi(20)C} \] Solving for \( C \), we get:
\[ C = \frac{1}{20\pi\sqrt{3} \times 100} \] Simplifying further:
\[ C = 10^{-4} \, \text{F} \]

Final Answer:
The capacitance \( C \) is \( 10^{-4} \, \text{F} = 100 \, \mu\text{F} \).

Answer 2

Step 1: Power in terms of voltage and resistance
The formula for power is given by:
\( P = \frac{V^2}{2} \)
We are given that the power \( P = 500 \) W, so:
\( 500 = \frac{100^2}{R} \)
Solving for \( R \), we get:
\( R = 20 \, \Omega \)

Step 2: Current calculation
Using Ohm's law across the resistor, we know:
\( P = I_{\text{rms}} \times V_{\text{rms}} \)
Given that \( P = 500 \) W and \( V_{\text{rms}} = 100 \) V, we can calculate the current \( I_{\text{rms}} \):
\( 500 = I_{\text{rms}} \times 100 \)
Solving for \( I_{\text{rms}} \), we get:
\( I_{\text{rms}} = 5 \, \text{A} \)

Step 3: Voltage and real power comparison
The RMS voltage across the resistance is given as \( V_{\text{rms}} = 200 \) V.
The real voltage (effective voltage) is 100 V, so the ratio is:
\( \frac{V_{\text{rms}}}{\text{real}} = 100 \, \text{V} \)

Step 4: Calculating the phase angle \( \varphi \)
The power factor \( \cos \varphi \) is the ratio of the real voltage to the RMS voltage:
\( \cos \varphi = \frac{100}{200} = \frac{1}{2} \)
Thus, the phase angle \( \varphi \) is:
\( \varphi = 60^\circ \)

Final Answer:
The phase angle \( \varphi \) is \( 60^\circ \).