Problems on AC to DC

Problem 1: A 200 mH coil is connected to an AC circuit with a 5 mA current. If the frequency is 2000 Hz then find out the voltage.Hz,

Solution:

Given:

L = 200 × 10^-3H

I = 5 × 10^-3A

f = 2000 Hz

We know that Inductive Reactance, X_L= L × ω = L × 2πƒ

= 2 × 3.14 × 2000 × 0.2

= 2512 Ohm

V = I X_L

= 0.005 × 2512

= 12.56 V

So, the voltage is 12.56 V

Problem 2: Find the instantaneous value of alternating voltage v = 5 sin(2π ×10⁴t) volt at:

0 s
20 μs
40 μs.

Solution:

At t = 0 s

v = 5 sin(0) = 0 V

At t = 20 μs

v = 5 sin(2π ×10⁴ × 20 × 10^-6)

= 5 sin( 40π × 10^-2)

= 5 sin(1.25)

= 5 × 0.94

= 4.74 V

At t = 40 μs

v = 5 sin(2π ×10⁴ × 40 × 10^-6)

= 5 sin( 80π × 10^-2 )

= 5 sin(2.5)

= 5 × 0.59 = 3 V

So, the instantaneous value of alternating voltage at 0 s, 20 μs, 40 μs is 0V, 4.74V, 3V respectively.

Problem 3: The current across an inductive coil is given by 0.7 sin (300t – 40°) A. Write the equation for the voltage if the inductance is 60 mH.

Solution:

L = 60 × 10^-3 H

i = 0.7 sin (300t – 40°) A

X_L = ωL = 300 × 60 × 10^-3 = 18 Ω

V_m = I_m XL = 0.7 × 18 = 12.6 V

In an inductive circuit, the voltage leads the current by 90° Therefore,

v = V_m sin ( ωt + 90°)

v = 12.6 sin(300t −40 + 90°)

v = 12.6 sin(300t +50º) V

So, the equation for voltage is v = 12.6 sin(300t +50°).

Problem 4: If the equation for an alternating current is given by i = 45 sin 314t. Then find the peak value, frequency, time period, and instantaneous value at t = 1 ms.

Solution:

i = 45 sin 314t; t = 1 ms = 1 × 10^-3 s

Comparing with the general equation of an alternating current, i = I_m sin ωt.

Peak value, I_m = 45 A

Frequency, f = ω/2π = 314 / 2 × 3.14 = 50 Hz

Time period, T = 1/f = 150 = 0 .02 s

At t = 2 ms,

Instantaneous value,

i = 45sin(3.14 × 1 × 10⁻³)

i = 0.14 A

Problem 5: If the DC voltage of the supply is 5V and the current is 2.5 A. Then find the resistance in the circuit.

Solution:

R = V/I

= 5/2.5

= 2 Ohm.

So, the resistance in the circuit is 2 Ohm.

Problem 6: Write down the equation for a sinusoidal voltage of 30 Hz and its peak value is 50V. Also, find time for one cycle.

Solution:

f = 30 Hz, V_m = 50V

v = V_msinωt

= V_m sin2πft

= 50 sin(2π × 30)t

= 50 sin(60 × 3.14)t

v = 50.sin(188t)

T = 1 / f = 1 / 30

= 0.033 sec = 33 ms

So, equation is v = 50 sin188t and time for one cycle 33 ms.

How to convert from AC to DC?

Electricity is a fundamental aspect of physics that deals with the behaviour and movement of electrically charged particles. When these charges remain stationary on a material’s surface, we refer to it as static electricity. However, when charges move, they create magnetic fields, and changes in magnetic fields can, in turn, generate electricity. This interplay between electric and magnetic fields is studied under the branch of physics known as electromagnetism.

Table of Content

Alternating Current (AC) Definition
Direct Current (DC) Definition
AC and DC Circuits

When do you need to convert DC to AC Power?