Solving for AC Voltmeters with a Pure Resistance and Pure Inductance

In summary: Knowing the magnitude and direction of the real and reactive voltages will help determine the appropriate resistor value.
  • #1
bendaddy
4
0

Homework Statement



a) A pure resistance and a pure inductance are connected in series across a 100 V(rms) AC line. An AC voltmeter gives the same reading when connected across either the R or the L. What does it read? (Note: Meters read rms values.)

b) The magnitudes of R and L are altered so that a voltmeter across L reads 50 V. What will the voltmeter read across R?


Homework Equations





The Attempt at a Solution


Solution should be: a) 70.7V ; b) 86.7V
 
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  • #2
The voltage across the line depends on a number of things:
1) the position of the input waveform in terms of phase
2) the total impedance of the line

The total impedance of the line is the sum of the resistance and the inductance.

You know that the average voltage on the line should be 100V under no load conditions.

The voltage dropped across the resistor is a constant. It depends on the resistance of the resistor, as the current through the circuit is a fixed RMS current (same current flows through resistor, inductor and to/from source).

the voltage dropped across the inductor depends on the rate of change of the current passing through it:
v(t) = L (di/dt)

So... we can say that the rate of change is pretty constant... as the frequency is constant... and because the question tells you that the two voltages are the same you can relate the inductance of the inductor and the resistance of the resistor.

V1 = IR V2 = L(Di/dt)

For the two things to have the same voltage... in terms of voltage divider theory.. they must have the same overall impedance...

so IR = L(di/dt)

You need to have values of R and L which provide the same numbers for all values of I.

Lets consider I as a function of time I = sin(2*Pi*f*t)R*cos(2*Pi*f*t) = L d(sin(2*Pi*f*t))/dt

we know that d/dx (sin (t) ) = cos(t)

So...

R*cos(2*Pi*f*t) = L*(2*Pi*f) cos(2*Pi*f*t))

now we can divide...

R = L (2*Pi*f)

I guess f must be 1/2*Pi

for R = L

Not sure how to calculate an exact R or L without knowing the frequency...
but if you knew the frequency and the R and the L you can find the voltages across them (which we know to be the same) without a problem.

Hope this helps.
Are you sure there are not more variables or given/known data??

Good Luck :)
 
  • #3
bendaddy said:

Homework Statement



a) A pure resistance and a pure inductance are connected in series across a 100 V(rms) AC line. An AC voltmeter gives the same reading when connected across either the R or the L. What does it read? (Note: Meters read rms values.)

b) The magnitudes of R and L are altered so that a voltmeter across L reads 50 V. What will the voltmeter read across R?


Homework Equations





The Attempt at a Solution


Solution should be: a) 70.7V ; b) 86.7V

You should provide an attempt at a solution, if only your own thoughts about how to approach the problem.

As a suggestion, investigate the voltage triangle for the circuit (similar to a power triangle, but places the real and reactive voltages on the "legs" of the triangle, with the hypotenuse representing the applied, or "apparent" voltage). This is analogous to a phasor analysis of the circuit.

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Related to Solving for AC Voltmeters with a Pure Resistance and Pure Inductance

What is an AC voltmeter?

An AC voltmeter is a device used to measure the voltage of an alternating current (AC) circuit. It typically consists of a needle or digital display that shows the magnitude of the voltage.

What is a pure resistance?

A pure resistance is a component in an electrical circuit that resists the flow of current without storing or releasing energy. It is represented by the symbol "R" and is measured in ohms (Ω).

What is a pure inductance?

A pure inductance is a component in an electrical circuit that stores and releases energy in the form of a magnetic field. It is represented by the symbol "L" and is measured in henries (H).

How do you solve for AC voltmeters with a pure resistance and pure inductance?

To solve for an AC voltmeter with a pure resistance and pure inductance, you can use Ohm's law and the formula V = IZ, where V is the voltage, I is the current, and Z is the impedance, which is the combination of the resistance and inductance. You can also use a phasor diagram to visualize the relationship between the voltage, current, and impedance.

What are some applications of solving for AC voltmeters with a pure resistance and pure inductance?

Solving for AC voltmeters with a pure resistance and pure inductance is important in many industries, such as electrical engineering, telecommunications, and power generation. It is also used in everyday devices, such as home appliances and electronic devices, to ensure the proper functioning of AC circuits.

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