Anonymous
more than 1 week ago
A 60 volt, 10 watt lamp is to be on 100 volt,60HZ A .C mains (I) calculate inductance of choke coil required (Ii) if a resistor is used in (I)?
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ChatGPT
more than 1 week ago
(I) To calculate the inductance of the choke coil required, we can use the formula:
XL = 2πfL
Where:
XL = inductive reactance
f = frequency of the AC mains (60 Hz)
L = inductance of the choke coil
Given that the lamp is 10 watts and the voltage is 60 volts, we can calculate the current flowing through the lamp:
P = VI
10 = 60I
I = 10/60
I = 0.167 A
Now, we can calculate the impedance of the lamp:
Z = V/I
Z = 60/0.167
Z = 359.28 ohms
Since the impedance of the lamp is equal to the inductive reactance of the choke coil, we can substitute the values into the formula:
XL = 2πfL
359.28 = 2π*60*L
L = 359.28 / (2π*60)
L ≈ 0.95 H
Therefore, the inductance of the choke coil required is approximately 0.95 H.
(ii) If a resistor is used in series with the choke coil, the total impedance will be the sum of the resistive and inductive components:
Ztotal = √(R^2 + XL^2)
Given that the resistor is in series with the choke coil, the total impedance should be equal to the impedance of the lamp (359.28 ohms). Therefore, we can set up the equation:
Ztotal = √(R^2 + XL^2)
359.28 = √(R^2 + (2π*60*L)^2)
359.28 = √(R^2 + (2π*60*0.95)^2)
Solving for R, we get:
R = √(359.28^2 - (2π*60*0.95)^2)
R ≈ 358.93 ohms
Therefore, if a resistor is used in series with the choke coil, the resistance required is approximately 358.93 ohms.
XL = 2πfL
Where:
XL = inductive reactance
f = frequency of the AC mains (60 Hz)
L = inductance of the choke coil
Given that the lamp is 10 watts and the voltage is 60 volts, we can calculate the current flowing through the lamp:
P = VI
10 = 60I
I = 10/60
I = 0.167 A
Now, we can calculate the impedance of the lamp:
Z = V/I
Z = 60/0.167
Z = 359.28 ohms
Since the impedance of the lamp is equal to the inductive reactance of the choke coil, we can substitute the values into the formula:
XL = 2πfL
359.28 = 2π*60*L
L = 359.28 / (2π*60)
L ≈ 0.95 H
Therefore, the inductance of the choke coil required is approximately 0.95 H.
(ii) If a resistor is used in series with the choke coil, the total impedance will be the sum of the resistive and inductive components:
Ztotal = √(R^2 + XL^2)
Given that the resistor is in series with the choke coil, the total impedance should be equal to the impedance of the lamp (359.28 ohms). Therefore, we can set up the equation:
Ztotal = √(R^2 + XL^2)
359.28 = √(R^2 + (2π*60*L)^2)
359.28 = √(R^2 + (2π*60*0.95)^2)
Solving for R, we get:
R = √(359.28^2 - (2π*60*0.95)^2)
R ≈ 358.93 ohms
Therefore, if a resistor is used in series with the choke coil, the resistance required is approximately 358.93 ohms.
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Google Gemini AI
more than 1 week ago
(i) The impedance of the lamp is given by:
```
Z = V / I = 60 V / (10 W / 60 V) = 6 ohms
```
The inductive reactance of the choke coil is given by:
```
XL = Z - R = 6 ohms - 0 ohms = 6 ohms
```
The inductance of the choke coil is given by:
```
L = XL / (2 - pi - f) = 6 ohms / (2 - pi - 60 Hz) = 0.016 H
```
Therefore, the inductance of the choke coil required is 0.016 H.
(ii) If a resistor is used in place of the choke coil, the resistance of the resistor must be equal to the impedance of the lamp:
```
R = Z = 6 ohms
```
Therefore, if a resistor is used, the resistance of the resistor must be 6 ohms.
```
Z = V / I = 60 V / (10 W / 60 V) = 6 ohms
```
The inductive reactance of the choke coil is given by:
```
XL = Z - R = 6 ohms - 0 ohms = 6 ohms
```
The inductance of the choke coil is given by:
```
L = XL / (2 - pi - f) = 6 ohms / (2 - pi - 60 Hz) = 0.016 H
```
Therefore, the inductance of the choke coil required is 0.016 H.
(ii) If a resistor is used in place of the choke coil, the resistance of the resistor must be equal to the impedance of the lamp:
```
R = Z = 6 ohms
```
Therefore, if a resistor is used, the resistance of the resistor must be 6 ohms.
0
