Showing posts with label RL RC Transformer. Show all posts
Showing posts with label RL RC Transformer. Show all posts

Sunday, 8 April 2018

Resistor and Capacitor in AC Circuit

Let us look at the behaviour of capacitor in ac circuit.Capacitor will create opposition to current flow which is called as reactance. The reactance depends of frequency and calculated using the formula:
\[\large Xc= \frac{1}{2\pi fC}\]
In pure capacitive circuit also the voltage lags the current by 90°


In a R-C circuit the opposition to current comes from both the resistor and also from the capacitor. The combined opposition is called impedance (Z).

Analysis of R-C circuit
R1 causes opposition to current  = 5Ω
C1 causes opposition to current =\[\large Xc= \frac{1}{2\pi 60\times 100\times 10^{-6}} = 26.5\]

\[\large Z=\sqrt{R^{2}+X^{2}}\]
\[\large Z=\sqrt{5^{2}+26.5^{2}} = 26.96\]


Current in the circuit = V /Z = 100 /26.96 = 3.71A
We can find other parameters such as voltage drop across components, true power and reactive power.

Voltage drop across R
VR = IR = 3.71 x 5 = 18.55V

Voltage drop across C
V =IX = 3.71 x 26.5 = 98.32V

Voltage supplied ( or Total voltage) =Vector addition of VR and VC

\[\large V_{T}=\sqrt{V_{R}^{2}+V_{C}^{2}}\]
\[\large V_{T}=\sqrt{18.55^{2}+98.32^{2}} = 100\]

Try to calculate the true power and reactive power.




Saturday, 7 April 2018

Resistor and Inductor in AC circuit

Basic behaviour of passive components in AC

The three passive components in electricity are resistor, inductor and capacitor. All these components impede (resist) current flow in an ac circuit. Resistor imposes resistance while inductor  and capacitor impose  reactance. Both these quantities are measured  in ohms. Just think of reactance as resistance in inductor and capacitor. 
The current and voltage behaviour of these components are quite interesting. In a pure resistive circuit, the current and voltage are always in phase with each other. While in a pure  inductive circuit, the applied voltage is ahead of current by 90°. 


In pure resistive circuit
Image result for voltage and current relation in resistor in ac





In pure inductive circuit 

The applied voltage leads the current by 90°. The applied voltage comes from the AC power supply. 

There is also another type of voltage which produced inside inductor which is created whenever there is change in current and this is called  induced voltage. The induced voltage always opposes the applied( source) voltage. Thus it is  always 180° apart from the applied voltage.

Phasor diagram of V applied, Current I and V induced.


For understanding we can compare pure resistive and pure inductive in one phasor diagram like shown below,


What if we have an AC circuit that has both resistor and inductor in it? What will be the phase difference between the applied voltage and current in the circuit?
In a circuit that has both resistance and inductance, the source voltage and source current (total current) will differ by an angle between 0 and 90°. The angle depends on the ratio of the R and L value.

R-L Series in AC circuit

Lets analyze the circuit below and determine all the circuit parameters such as current, voltage drop across components and power.

Reactance by L1 :  XL = 2πf L = 2 x 3.142 x 50 x 0.01 = 3.142 Ω
Resistance by R1 = 5Ω
The total obstacles to current flow is called impedance ( Z) whereby Z is the vector addition of R and X.

Impedance diagram

Image result for impedance of R and L vector addition

In vector addition we cannot simply add the values directly unless there both quantities are in the same direction. In the R-L case we cannot just add the value of R and X even if they are in series.

\[\large Z= \sqrt{R^{2}+X^{2}}\]

Applying the formula in the analysis:


\[\large Z= \sqrt{5^{2}+3.142^{2}} = 5.90\]

I = V/Z
= 100/5.9 = 16.95A

Voltage drop across resistor
V = IR = 16.95 x 5 = 84.75V

Voltage drop across inductor
VL = IX = 16.95 x 3.142 = 53.26 V


When both voltages are added directly, the total becomes 84.75 + 53.26 = 138.0 V , which exceeds our supply voltage of 100V.
This is not possible. 

Again, remember that these quantities are not in the same phase. Vector addition must be done to find the total voltage or the supplied voltage and not normal addition. 

\[\large V_{T}= \sqrt{V_{R}^{2}+V_{X}^{2}}\]


\[\large V_{T}= \sqrt{84.75^{2}+ 53.26^{2}} = 100\]



Leading and Lagging self-test

From the waveform below, can you identify which waveform is in phase, which is leading and lagging??