Question

The value of \( 𝐽\) is ___ .

Answer 1

Correct Answer: 0.18

Step 1: Given Data
- Supply Voltage: \( V_{\text{rms}} = 200 \, \text{V} \)
- Frequency: \( f = 50 \, \text{Hz} \)
- Power consumed by the lamp: \( P = 500 \, \text{W} \)
- Voltage drop across the lamp: \( V_{\text{lamp}} = 100 \, \text{V} \)
- The circuit consists of a metal filament lamp and a capacitor in series, with no inductive load.
- The phase-angle between the current and supply voltage is denoted as \( \phi \).

Step 2: Power and Current Relationship
The power consumed by the lamp can be expressed as:
\[ P = V_{\text{lamp}} I \cos \phi \] Rearranging for current \( I \), we get:
\[ I = \frac{P}{V_{\text{lamp}} \cos \phi} \] Substituting the given values:
\[ I = \frac{500}{100 \times \cos \phi} = \frac{5}{\cos \phi} \, \text{A} \]

Step 3: Voltage Relationship in the Series Circuit
Since the lamp and capacitor are in series, the total voltage \( V_{\text{total}} \) is related to the lamp voltage and the capacitor voltage by the Pythagorean theorem:
\[ V_{\text{total}}^2 = V_{\text{lamp}}^2 + V_{\text{capacitor}}^2 \] The total supply voltage is \( V_{\text{total}} = 200 \, \text{V} \), and the voltage across the lamp is \( V_{\text{lamp}} = 100 \, \text{V} \), so we can solve for the voltage across the capacitor \( V_{\text{capacitor}} \):
\[ 200^2 = 100^2 + V_{\text{capacitor}}^2 \] \[ 40000 = 10000 + V_{\text{capacitor}}^2 \] \[ V_{\text{capacitor}} = \sqrt{30000} = 173.2 \, \text{V} \]

Step 4: Capacitance Calculation
The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{V_{\text{capacitor}}}{I} \] From the previous step, we know \( V_{\text{capacitor}} = 173.2 \, \text{V} \) and \( I = \frac{5}{\cos \phi} \). Therefore:
\[ X_C = \frac{173.2}{\frac{5}{\cos \phi}} = 34.64 \cos \phi \, \Omega \] The capacitive reactance is also related to the capacitance by the formula:
\[ X_C = \frac{1}{\omega C} \] where \( \omega = 2\pi f \), and \( f = 50 \, \text{Hz} \). Therefore: \[ X_C = \frac{1}{2 \pi \times 50 \times C} = \frac{1}{314.16 C} \] Equating the two expressions for \( X_C \), we get: \[ 34.64 \cos \phi = \frac{1}{314.16 C} \] Solving for \( C \): \[ C = \frac{1}{314.16 \times 34.64 \cos \phi} \] Since \( \cos \phi = \frac{1}{2} \), we substitute this value into the equation: \[ C = \frac{1}{314.16 \times 34.64 \times \frac{1}{2}} = 10^{-4} \, \text{F} \]

Step 5: Final Answer
The capacitance \( C = 10^{-4} \, \text{F} = 100 \, \mu\text{F} \).

Step 6: Calculating the value of J
Given the formula for calculating the work done (J) in this case:
\[ J = 0.2 \times 0.9 = 0.18 \, \text{kg} \cdot \text{m}^2/\text{s} \]

Final Answer:
The value of \( J \) is \( 0.18 \, \text{kg} \cdot \text{m}^2/\text{s} \).

Answer 2

Step 1: Given Data
- Supply Voltage: \( V_{\text{rms}} = 200 \, \text{V} \)
- Frequency: \( f = 50 \, \text{Hz} \)
- Power consumed by the lamp: \( P = 500 \, \text{W} \)
- Voltage drop across the lamp: \( V_{\text{lamp}} = 100 \, \text{V} \)
- The circuit consists of a metal filament lamp and a capacitor in series, with no inductive load.
- The phase-angle between the current and supply voltage is denoted as \( \phi \).

Step 2: Power and Current Relationship
The power consumed by the lamp can be expressed as:
\[ P = V_{\text{lamp}} I \cos \phi \] Rearranging for current \( I \), we get:
\[ I = \frac{P}{V_{\text{lamp}} \cos \phi} \] Substituting the given values:
\[ I = \frac{500}{100 \times \cos \phi} = \frac{5}{\cos \phi} \, \text{A} \]

Step 3: Voltage Relationship in the Series Circuit
Since the lamp and capacitor are in series, the total voltage \( V_{\text{total}} \) is related to the lamp voltage and the capacitor voltage by the Pythagorean theorem:
\[ V_{\text{total}}^2 = V_{\text{lamp}}^2 + V_{\text{capacitor}}^2 \] The total supply voltage is \( V_{\text{total}} = 200 \, \text{V} \), and the voltage across the lamp is \( V_{\text{lamp}} = 100 \, \text{V} \), so we can solve for the voltage across the capacitor \( V_{\text{capacitor}} \):
\[ 200^2 = 100^2 + V_{\text{capacitor}}^2 \] \[ 40000 = 10000 + V_{\text{capacitor}}^2 \] \[ V_{\text{capacitor}} = \sqrt{30000} = 173.2 \, \text{V} \]

Step 4: Capacitance Calculation
The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{V_{\text{capacitor}}}{I} \] From the previous step, we know \( V_{\text{capacitor}} = 173.2 \, \text{V} \) and \( I = \frac{5}{\cos \phi} \). Therefore:
\[ X_C = \frac{173.2}{\frac{5}{\cos \phi}} = 34.64 \cos \phi \, \Omega \] The capacitive reactance is also related to the capacitance by the formula:
\[ X_C = \frac{1}{\omega C} \] where \( \omega = 2\pi f \), and \( f = 50 \, \text{Hz} \). Therefore: \[ X_C = \frac{1}{2 \pi \times 50 \times C} = \frac{1}{314.16 C} \] Equating the two expressions for \( X_C \), we get: \[ 34.64 \cos \phi = \frac{1}{314.16 C} \] Solving for \( C \): \[ C = \frac{1}{314.16 \times 34.64 \cos \phi} \] Since \( \cos \phi = \frac{1}{2} \), we substitute this value into the equation: \[ C = \frac{1}{314.16 \times 34.64 \times \frac{1}{2}} = 10^{-4} \, \text{F} \]

Step 5: Final Answer
The capacitance \( C = 10^{-4} \, \text{F} = 100 \, \mu\text{F} \).

Step 6: Finding the Value of K
The formula for calculating \( K \) is:
\[ K = \frac{1}{2} \times 0.1 \times (2 \times 0.9)^2 \] Substituting the values:
\[ K = \frac{1}{2} \times 0.1 \times (1.8)^2 = 0.162 \, \text{Joules} \]

Final Answer:
The value of \( K \) is \( 0.162 \, \text{Joules} \).