Ohm`s law V=IR. So what is a Volt ? Potential energy per unit charge or work done to move a charge against electric field, form one potential to a higher potential.

Dynamic resistance R_{dynamic}= ΔV/ΔI it is the local resistance – the tangent to the slope of the device V-I curve.

Power P=I×V or derived: P=I²R or P=V²/R

Kirchhoff`s law: sum of voltages around circuit is zero, sum of currents in and out of node is zero.

Parallel resistance: R_{total} = (R1×R2)/(R1+R2)

Rule of thumb: if two parallel resistors differ by a factor of ten or more, then we can ignore the larger of two.

In other words: in a parallel circuit, resistor much smaller than other dominates. Ina series circuits, the larger resistor dominates.

Voltage divider. Rely on fact the I is the same on top and bottom.

V_{out} = V_{in} × R2 / (R1+R2)

Tip: since the current at the top and bottom are equal, the voltage drops are proportional to the resistances.

So if the lower leg is ten times the upper leg, it will show 90% of the input voltage ( or 10/11 if exactly).

Or if lower leg is 3 times than upper leg, output voltage should be 75% of input voltage ( or 3/4 if exactly ), lets check that:

Thevenin`s model. How to calculate circuit when it gets loaded? Model the actual circuit with the simpler circuit – Thevenin model – which is idealized voltage source with resistor in series.

Then we can rapidly see how that circuit behaves under various loads.

V_{thevenin} is just open circuit, the voltage when nothing is attached ( no load ).

R_{thevenin} is V_{thevenin}/I_{short-circuit }

The fastest way to calculate R_{thevenin} is to see it as the parallel resistance of several resistances viewed from the output.

Any non ideal voltage source droops when loaded. How much it droop depends on it output impedance. The Thevenin equivalent model, with its R_{thevenin} describes this property neatly in a single number.

Design rule of thumb: when circuit A drives circuit B. Let R_{out} for A be <= 1/10 R_{in} for B.

It means we will pass strong signal to B without big droop.

If R_{outA} is much smaller then R_{inB} then the divider delivers nearly the all of the original signal.

Labs.

IL.1 nothing to do here, just some information how to use lab stuff.

IL.2 Voltage divider. I have used 10k resistors for voltage divider and 3.9k for load. Results:

So we have voltage divider with 10k resistors. We calculate R_{thevenin} 5k and V_{thevenin} 7.5v.

We measure that V_{open} is 7.49V that confirms our calculations.

also we measure our I_{short-circuit} to be 1.5mA

After measuring we calculate R_{thevenin} and it is 4.9K and that confirms what we have calculated previously.

Now we load circuit with 3.9V and measure that voltage droop or V_{load} is 3.2V

And our calculation later confirms our measurements.

Now, if we would like to deliver almost all of the signal to load, by our rule of thumb we should choose R_{load} to be at least 10 times R_{thevenin. }Lets choose R_{load} 56k and see what happens.

And that confirms our rule of thumb: that when loaded with at least 10 times higher resistance we got 90 percent of our signal delivered.

1L.3 I have used analog VOM, in 10v scale I have measured 200k internal resistance.

So it means it moves full scale on 10v with its internal resistance of 200k.

How could one use it to measure for example 50v or 100k ?

We must divide this voltage so that on 50v for example scale would behave same like on 10v, and same for 100v.

We have to add two voltage dividers and calculate resistor values.

We can calculate resistance in two ways: algebraic or just talk our way out.

The bigger the resistance, the more voltage we lose in a dividers leg. In case of 50v we now that we must have 1/5 (10v) on lower leg. Se we must loose 4 from 5 in upper leg. We must choose proportionally bigger resistor in upper leg. 200K times 4 is 800k. That is the value for upper leg resistor in case of 50v.

1L3.3 To get 10mA ammeter from voltmeter we fist calculate current when itsi reading maximum 10v using its internal 200k resistance.

It is I_{0}=10v/200k=50uA

10mA that we need to measure is 200 time bigger that 50uA

So we would add parallel resistor to take 199 parts of current fro 200k and we get 1k resistor. As 1k is way higher that 200k, our internal resistance becomes 1k.

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