Now let's take a break for another Ohm's law and Kirchhoff's laws practice problem.
This time we want to find:
Value of R3
Value of R5
Voltage at TP1
Voltage at TP2
Voltage at TP3
This circuit is a little more complicated so let's break it down to its parts before moving on.
Here the left part of the circuit is
highlighted. This part of the circuit is in series; there is only one
current path. The current is the same everywhere (2mA). The value of R1
is twice the value of R6 so the voltage across R1 is twice the voltage
across R6. Be careful, the 45 volts of the battery is not shared by
these resistors (we don't have 30 volts across R1 and 15 volts across
R6). There is more to this circuit.
The right part of the circuit is in
parallel; there are multiple current paths. The voltage will be the
same across the resistors and the current is split up inversely
proportional to the resistors (the greater resistance has the lower
current).
If we zoom out, R4 and R5 can be seen
as a single resistance in series with R3. We don't know what value this
resistance is, but we know that the two legs of 500 microamps of
current each will combine to make 1 milliamp through the combined
resistance. Since R3 is in Series with R4/R5 it will also have this 1
mA of current flowing through it.
Now R3 can be added to R4/R5 to make
a single resistance. This resistance is in parallel with R2. Now TP2 is
embedded somewhere in the combined resistance so we can't show it.
Of course, we can now combine R3/R4/R5 with R2 but I think you get the
drift. Now that we have looked at the parts of the circuit, let's solve
the puzzle.
The voltage at TP1 is ready to solve. We start with 45 volts, lose some
across R1 and whatever is left is the voltage at TP1. Let's calculate
it. We know the resistance and the current. We are trying to find the
voltage so we multiply 2 mA times 10k. That's 0.002 amps times 10,000
ohms...
.
0
0
2
X
1
0
0
0
0
=
20
...for 20 volts across R1. Subtract that from the battery voltage of 45 volts and we have 25 volts at TP1.
We can also find the voltage at TP3.
First we need the voltage across R6. We can find this one of two ways.
We can use Ohm's law: 2mA times 5k. On the calculator that's...
.
0
0
2
X
5
0
0
0
=
10
...for 10 volts. We can also see that
R1 and R6 are in series with each other. R6 has half the resistance of
R1. Being in series with R1, R6 must have half the voltage, which is 10
volts. Now, how do we use the voltage across R6 to find the voltage at
TP3? First, the negative side of the battery is 0 volts; it is ground.
Conventional current is traveling toward the negative side of the
battery (right to left). The voltage at TP3 must be higher than the
voltage on the left side of R6 (the voltage is higher where
conventional current enters a resistor). We have 0 volts on the left
side of R6 and 10 more volts on the right side. TP3 has +10 volts.
There is a 10 volt drop across R6 (R6 has half the resistance of R1 and
they are in series with each other). Going from left to right that is
actually a gain of 10V. Starting with 0 volts on the negative side of
the battery this gives us +10 volts at TP3.
Now we have enough information to
calculate the voltage at TP2. The bottom of R4 will have the same
voltage as TP3. If we add the voltage across R4 we will have the
voltage at TP2. Let's calculate the voltage across R4. We have 500μA
through 20k. That's 0.0005 amps through 20,000 ohms. On the calculator
that's...
.
0
0
0
5
X
2
0
0
0
0
=
10
...for 10 volts. Add this to the 10 volts at the bottom of the resistor and we have 20 volts at TP2.
The bottom of R4 has the same voltage as TP3 (+10 volts). R4 has a
voltage drop of 10 volts (500 microamps x 20 kiloohms). This is
actually a gain of 10 volts going from the bottom to the top. This
gives us +20 volts at TP2.
Now for the value of R3. The voltage
across R3 is the difference between the voltages at TP1 and TP2. That's
5 volts. Now what's the current. The current, going from left to right
through R3 splits and goes through R4 and R5, each getting 500μA.
Kirchhoff's Current Law says we add these together to see what went
through R3. That's 1,000μA or 1mA. We divide that into our 5 volts.
That's 5 volts divided by 1mA. On the calculator...
5
.
0
0
1
=
5000
...for 5,000 ohms or 5k for R3.
R3, R4 and R5 constitute a series/parallel circuit. We don't know the
current through R3 but it must be the sum of the currents through R4
and R5. R4 and R5 each have 0.5 milliamp (500 microamps) so the current
through R3 is 1 milliamp. The voltage across R3 is the difference
between TP1 and TP2. That's 5 volts. 5 volts divided by 1 milliamp
gives us 5k for R3.
R5 and R4 are in parallel, so they
have the same voltage, which we calculated above as 10 volts. 10 volts
divided by 500μA on the calculator is...
1
0
.
0
0
0
5
=
2000
...for 20,000 ohms or 20k for R5. Actually, a quick look will show that R4 and R5 have the same voltage and the same current so they must be the same value.
R4 and R5 have the same current. They are in parallel with each other
so they have the same voltage. We know the current is 500 microamps and
the voltage is 10 volts. However, just knowing that the voltages and
currents are equal tells us that the resistors are of equal value. R4
is 20k and so is R5.
Just for the heck of it, let's prove
that R2 is 15k. The voltage across R2 is the difference between the
voltages a TP1 and TP3. That's 15 volts. What is the current through
R2? We started with 2mA and that splits off between R3 and R2. We
already determined that R3 has 1mA flowing through it, so the other
milliamp must be flowing through R2 (Kirchhoff's Current Law). We have
15 volts and 1mA: 15 .001 = 15000 for 15,000 ohms or 15k for R2.
The current through R1 (2 mA) splits two ways through R2 and R3. We
have 1 mA through R3 so the other milliamp must go through R2. The
voltage across R2 is the difference between TP1 and TP3. That's 15
volts. Divide that by 1 milliamp and we get 15k, just as the schematic
says.
