A pure inductance of 1.273mh is connected with series with a pure resistance of 50 ohms if the frequency of the sinusoidal supply is 5khz and the pd across the 30 ohms is 6
1, Draw a circuit diagram
2, Determine the value of the supply voltage
3, Fined the voltage across the insuctance
4, Draw a phasore diagram
1. Circuit diagram:
```
+--L(1.273mH)--R(50 ohms)--+
| |
V |
| |
+-----------(6V)------------+
```
2. Value of supply voltage:
Given the resistance R = 50 ohms, and the PD across it = 6V
Using Ohm's Law: V = IR
Supply voltage = V + IR = 6V + (50 ohms)(6V) = 6V + 300V = 306V
3. Voltage across the inductance:
Inductive reactance, XL = 2πfL
= 2 * π * 5000 Hz * 1.273mH
= 10π ohms
= 31.732 ohms
Voltage across inductance = IXL = IX × 31.732 V
To find:
Current, I = V / (R + XL)
= 306 / (50 + 31.732)
≈ 306 / 81.732
≈ 3.74 A
Thus, voltage across inductance ≈ 3.74 A * 31.732 V
≈ 118.8 V
4. Phasor diagram:
```
V
/ |
/ |
/ |
+----+--------- I * R = 6 V
| |
| | In L
| /
| /
| /
| /
+------+------- Supply voltage
```
In the phasor diagram, V represents the supply voltage, I*R represents the voltage across the resistance, and In*L represents the voltage across the inductance.