Solving Series LR Circuit: Calculate Wattless & Power Components

In summary, the conversation discusses the calculation of the wattless and power components, as well as the total current, in a series circuit consisting of a self inductance coil and a non-inductive resistance connected to a 200-volt supply at a frequency of 50 cycles per second. The concept of complex impedance and its relationship to the power and wattless components is also mentioned.
  • #1
Amith2006
427
2

Homework Statement



1)A coil of self inductance of 0.7 Henry is joined in series with a non inductive resistance of 5 ohms. Calculate the wattless and power components as well as total current when connected to a supply of 200 volts at frequency of 50 cycles per second.

Homework Equations





The Attempt at a Solution



It is given in my book that the power component of current = I(1)[cos(theta)]
Where I(1) = Root Mean Square value of total current in the circuit
I think theta is the phase difference between current and voltage in circuit
Also they say that the wattless component of current = I(1)[sin(theta)]
Could somebody please explain, what is the concept behind this?
 
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  • #2
I'm not familiar with those exact terms, but I think I can see what they are driving at.

Try this -- plot the complex impedance vector in the complex plane, with the real (resistive) axis on the horizontal and the imaginary (reactive) axis on the vertical. Calculate the complex impedance looking into the series combination of the L and R, and plot that as a vector in this 2-space. The horizontal component is just the resistance, and the vertical component is the reactive impedance of the inductor at that frequency. So if you have the total power as P = |Z| I(1), then it would make sense what they are saying about the cosine component being the portion across the resistor. Does that work out?
 
  • #3
berkeman said:
I'm not familiar with those exact terms, but I think I can see what they are driving at.

Try this -- plot the complex impedance vector in the complex plane, with the real (resistive) axis on the horizontal and the imaginary (reactive) axis on the vertical. Calculate the complex impedance looking into the series combination of the L and R, and plot that as a vector in this 2-space. The horizontal component is just the resistance, and the vertical component is the reactive impedance of the inductor at that frequency. So if you have the total power as P = |Z| I(1), then it would make sense what they are saying about the cosine component being the portion across the resistor. Does that work out?

I got your point.Thanks buddy.:smile:
 

FAQ: Solving Series LR Circuit: Calculate Wattless & Power Components

What is a series LR circuit?

A series LR circuit is a circuit that contains a resistor (R) and an inductor (L) connected in series. The resistor limits the current flow, while the inductor stores energy in the form of a magnetic field. This type of circuit is commonly used in AC circuits.

How do you calculate wattless components in a series LR circuit?

The wattless component in a series LR circuit is the reactive power, also known as the wattless power or the volt-ampere reactive (VAR). It is calculated by multiplying the voltage (V) by the inductive reactance (XL) and dividing by the total impedance (Z). The formula is: VAR = (V * XL) / Z.

How do you calculate power components in a series LR circuit?

The power components in a series LR circuit are the active power, also known as the real power or the watt (W), and the reactive power (VAR). The active power is calculated by multiplying the voltage (V) by the current (I) and the power factor (PF). The reactive power is calculated using the formula mentioned in the previous question.

What is the power factor in a series LR circuit?

The power factor in a series LR circuit is a measure of how efficiently the circuit is using power. It is the ratio of the active power (W) to the apparent power (VA). A power factor of 1 indicates that all the power is being used effectively, while a power factor less than 1 indicates that some power is being wasted due to the presence of reactive components.

How do you solve a series LR circuit?

To solve a series LR circuit, you need to follow these steps:
1. Calculate the total impedance (Z) by adding the resistive and inductive components in a vector form.
2. Calculate the current (I) by dividing the voltage (V) by the total impedance (Z).
3. Calculate the active power (W) by multiplying the voltage (V) by the current (I) and the power factor (PF).
4. Calculate the reactive power (VAR) using the formula mentioned in question 2.
5. Calculate the apparent power (VA) by dividing the active power (W) by the power factor (PF).
6. Use the values obtained to draw a phasor diagram and analyze the circuit.

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