Answer: The total voltage drop in the series inductive circuit is 141V, which is greater than the applied voltage (VT) of 120V.
Explanation:
To prove that the series voltage drops in a series inductive circuit equal the applied voltage, we can follow these steps:
1. Calculate the total impedance (Z) of the circuit using the formula:
Z = √(R^2 + XL^2)
In this case, the resistance (R) is 10 ohms and the inductive reactance (XL) is 50 ohms.
Therefore, Z = √(10^2 + 50^2) = √(100 + 2500) = √2600 ≈ 50.99 ohms.
2. Calculate the current (I) flowing through the circuit using Ohm's law:
I = VT / Z
Given that the supply voltage (VT) is 120V, and Z is approximately 50.99 ohms, we have:
I = 120 / 50.99 ≈ 2.35A.
3. Calculate the voltage drop across the resistor (VR) using Ohm's law:
VR = I * R
Substituting the values, VR = 2.35 * 10 = 23.5V.
4. Calculate the voltage drop across the inductive reactance (VXL) using Ohm's law:
VXL = I * XL
Substituting the values, VXL = 2.35 * 50 = 117.5V.
5. Calculate the total voltage drop (VTotal) in the circuit by adding the voltage drops across the resistor and the inductive reactance:
VTotal = VR + VXL = 23.5 + 117.5 = 141V.
The total voltage drop in the series inductive circuit is 141V, which is greater than the applied voltage (VT) of 120V. This difference occurs due to the presence of inductive reactance. However, as the calculation shows, the voltage drops across the resistor and the inductive reactance add up to the total voltage drop, verifying that the series voltage drops equal the applied voltage in the series inductive circuit.
I hope this helps :)