Question

When a direct current of 2 A is passed through a coil, the potential difference across the coil
is 20 V. When an alternating current of 2 A at frequency 40 Hz is passed through a coil, the
potential difference across the coil is 140 V. Find the current in the coil if it is connected to
a 230 V, 50 Hz supply.

Answer 1

Answer on Question #52418, Engineering, Other

When a direct current of 2 A is passed through a coil, the potential difference across the coil is 20 V. When an alternating current of 2 A at frequency 40 Hz is passed through a coil, the potential difference across the coil is 140 V. Find the current in the coil if it is connected to a 230 V, 50 Hz supply.

Solution:

From direct current we can obtain resistance:

$$R = \frac{V}{I} = \frac{20}{2} = 10\ \Omega$$

Inductive reactance

$$X_L = 2\pi f L$$

Impedance

$$Z = \sqrt{R^2 + X_L^2}$$

Current

$$I_1 = \frac{V_1}{Z_1}$$

Thus, from given

$$Z_1 = \frac{V_1}{I_1} = \frac{140}{2} = 70\ \Omega$$

Hence,

$$X_L = \sqrt{Z_1^2 - R^2} = \sqrt{70^2 - 10^2} = 40\sqrt{3}\ \Omega$$

The Inductance

$$L = \frac{X_L}{2\pi f_1} = \frac{40 * \sqrt{3}}{2 * \pi * 40} = \frac{\sqrt{3}}{2\pi}$$

Thus, for 230 V, 50 Hz supply we have

$$I = \frac{V_2}{Z_2} = \frac{V_2}{\sqrt{R^2 + (2\pi f_2 L)^2}} = \frac{230}{\sqrt{10^2 + \left(2 * \pi * 50 * \frac{\sqrt{3}}{2 * \pi}\right)^2}} = \frac{230}{\sqrt{10^2 + \left(50 * \sqrt{3}\right)^2}} = 2.64\ \text{A}$$

Answer: 2.64 A

```cpp

#include <iostream>

#include <cstdlib>

using namespace std;

int main() {

int number;

cout << "Enter the number from 1 to 10 : ";

cin >> number;

if (cin.fail() || number < 1 || number > 10) {

cout << "wrong input\n";

system("pause");

}

}</cstdlib></iostream>