Ohm's Law & AC Homework Solutions

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

I believe none of the options are correct .

Option c) would have been correct if E represented peak voltage , I represented peak current and Z is used instead of R .

But I suppose the question is considering E to be E(t) and I to be I(t) .

Am I right ?

 

[gneill]

For AC circuits, (c) would apply whether you use peak or rms values, or whether using time domain or phasor values (you may or may not have been introduced to phasors at this point).

 

[Chandra Prayaga]

Why do you think none of the options are correct?

 

[Jahnavi]

For AC circuits, (c) would apply whether you use peak or rms values, or whether using time domain or phasor values (you may or may not have been introduced to phasors at this point).

If E(t) represents instantaneous voltage and I (t) represents instantaneous current, then E(t) ≠ I(t) Z , because E and I might not be in phase .

 

[gneill]

If E(t) represents instantaneous voltage and I (t) represents instantaneous current, then E(t) ≠ I(t) Z , because E and I might not be in phase .

Instantaneous values have no phase. Ohm's law applies to instantaneous values.

 

[Jahnavi]

Instantaneous values have no phase. Ohm's law applies to instantaneous values.

Please remove a confusion .

Consider an LR series circuit with applied source voltage E . E leads the current by some phase Φ . Expression of E is E = E_osin(ωt+Φ) . Z is the impedance .

But we cannot say that expression of current I = (1/Z)E_osin(ωt+Φ) .

Does this not mean that Ohm's law is not valid ?

 

[gneill]

Please remove a confusion .

Consider an LR series circuit with applied source voltage E . E leads the current by some phase Φ . Expression of E is E = E_osin(ωt+Φ) . Z is the impedance .

But we cannot say that expression of current I = (1/Z)E_osin(ωt+Φ) .

Does this not mean that Ohm's law is not valid ?

No. Z is an impedance, so it is a complex value. It can be written as a magnitude and phase. Perhaps you're confusing impedance with reactance?

 

[Jahnavi]

My confusion is that we can express peak applied voltage as a product of impedance and peak current in the circuit i.e E₀ = I₀Z ,satisfying Ohm's law .

But we cannot express E as product of Z and I₀sinωt ?

 

[gneill]

My confusion is that we can express peak applied voltage as a product of impedance and peak current in the circuit i.e E₀ = I₀Z ,satisfying Ohm's law .

But we cannot express E as product of Z and I₀sinωt ?

Z is a complex value, good for all time, but it is a phasor constant that doesn't, in general, work and play well with time-domain functions without some careful thought. I₀sinωt yields an instantaneous value at particular times. You could write what you did, with the understanding that you would obtain a function E(t) giving an instantaneous, complex value for the voltage. That complex value might be interpreted as voltage magnitude and phase with respect to the current. The issue then becomes one of interpreting the results in a useful manner. But nevertheless, Ohm's law will hold.

In practice we tend to use impedances in combination with phasor values for voltage and current, using some particular voltage supply to set the zero phase reference. But you can mix impedances with time domain functions if you're careful about the interpretation of the results.

If you think about instantaneous values only, whether peak or rms values, then the AC voltage will be described by some function E(t), and the current by some function I(t). Then for any instant in time there will be some value of resistance R(t) = E(t)/I(t). But this can give you the problematical situation where the resistance R(t) is not a constant over time. But, if you write the voltage and current as phasor values, say $E = E_o \angle \phi$ and $I = I_o \angle \alpha$, then you can write $Z = E/I$ and Z will be a constant (complex) value.

 

[Jahnavi]

Thanks gneill !

 

[cnh1995]

If E(t) represents instantaneous voltage and I (t) represents instantaneous current, then E(t) ≠ I(t) Z , because E and I might not be in phase

This is correct, but as gneill said, instantaneous values don't have phase.
Consider an RL circuit excited with an ac source. Since the current lags the applied voltage, you get instants where the applied voltage is non-zero but the current is zero. If E(t)=I(t)*Z applies here, Z becomes infinite.
So when you say E=IZ, E and I are peak (or rms) values. If you mention the phases of E, I and Z, they become phasors.
If you want to use E(t) and I(t), you need to use the v-i relationships of the components and apply KVL to solve for current.

 

[Delta2]

if we view V(t) and I(t) as complex valued functions of the real variable t, then the ratio $Z=\frac{V(t)}{I(t)}$ is a well defined complex number independent of time. So it holds for the instantaneous complex values that $V(t)=ZI(t)$ but it doesn't necessarily hold for the instantaneous real values. Z is here a complex number not to be confused with $|Z|$ which is the magnitude (real number) of Z and used for the peak values $V_0=|Z|I_0$

https://en.wikipedia.org/wiki/Electrical_impedance#Complex_voltage_and_current