Ohm's Law and Kirchhoff's Law Exercises

## Exercise, Ohm's law and Kirchhoff's law practice:

Understanding Ohm's law and Kirchhoff's laws is essential to analyzing any electronic circuit. Let's take this first of several opportunities to practice Ohm's law and Kirchhoff's laws to be sure we thoroughly understand them.

In the following circuit, TP1 and TP2 are test points where we measure voltage. Part of this exercise is to calculate the expected voltages at these points. We will make the typical assumption that the designated ground is the negative terminal of the battery. That is where the black probe of the imaginary volt meter will be placed. The other unknowns are the values of R1 and R3. Take all the time you need to solve this puzzle before reading the solution. Find:
R1  ____ Ω
R3  ____ Ω
TP1 ____ V
TP2 ____ V

### Solution:

First look for components where you already have enough information to use Ohm's law to calculate everything about that component. Let's start with R2. R2 has a value of 1.5 megaohms and has 300 microamps of current flowing through it. With this we can calculate the voltage across it. Get your calculator and follow along. We have 1.5 Mega-something, so we move the decimal point six places to the right for 1,500,000. This gives us the resistance in ohms. We also have 300 micro-something, so we move the decimal point six places to the left for 0.000,3. This gives us the current in amps. We don't know the voltage so we multiply the current by the resistance. On the calculator that is...

 1 5 0 0 0 0 0 X . 0 0 0 3 = 450

That will be 450 volts. Since the bottom of R2 is connected to ground (the current meter is essentially a short circuit so there is no voltage across it) and the voltage at TP1 is measured in relation to ground. The voltage across R2 determines the voltage at TP1, which is 450V. Now let's do the same thing for R4. R4 has a value of 500 kiloohms and has 600 microamps of current flowing through it. We have 500 kilo-something, so we move the decimal point three places to the right for 500,000 ohms. We also have 600 micro-something, so we move the decimal point six places to the left for 0.000,6 amps. Now we multiply again. On the calculator that is...

 5 0 0 0 0 0 X . 0 0 0 6 = 300

Giving 300 volts. Again, R4 is connected to ground so the voltage across it is also the voltage at TP2, which is 300V.

Now we know the voltages at both TP1 and TP2. Since voltage is always a difference we now know the voltage across R3. If only we knew the current through R3 we could calculate its value. Actually, we do know the current through R3. R3 and R4 are in series with each other (only one current path). Whatever current goes through R4 must first go through R3 (so to speak). Therefore, R3 has 150 volts across it and 600 microamps of current going through it. Let's calculate its value. We know the voltage so we divide into it...

 1 5 0 . 0 0 0 6 = 250000

Move the decimal point three places to the left and that makes R3 250 k.

Now let's tackle R1 We start out with 540 volts and end up with 450 volts after passing R1. There is a drop of 90 volts across R1. Now all we need is the current through R1 to calculate its value. After R1 the circuit changes from a series circuit to a parallel circuit. Kirchhoff's current law says that the currents in the two legs of the parallel circuit must add up to the total current that started from the battery. This means that R1 must have 900 microamps flowing through it. Now we just divide that into the 90 volts across it and we have its value. On the calculator that is...

 9 0 . 0 0 0 9 = 100000

R1 has a value of 100,000 ohms. Move the decimal point three places to the left and we have 100k.

The final answers are:

R1:  100k
R3:  250k
TP1: 450V
TP2: 300V