62 GEARS September 2005
and
Kirchoff’s Laws of Current and Voltage
So far in our Pushing Electronsseries we’ve looked at what makes electricity; how to
define electricity using common terms; and Ohm’s Law, and how it predicts the characteristics of electricity in a circuit. Maybe its time we stop talking about it, and start testing those characteristics and rules.
We’ll begin our test with a simple circuit: just a single load in the circuit. As we learned earlier, every circuit must include these three basic compo-nents: • a power supply• a load• conductors
There are certain principles that are common to all simple circuits. One might consider these to be the laws of simple circuits: 1. The circuit will use all of the volt-age available to it.2. The voltage will be the same through-out the power side of the circuit.3. The voltage will be the same through-out the ground side of the circuit.
4. Amperage will be the same through-out the circuit.5. The total resistance of the circuit will be equal to the sum of each com-ponent’s resistance.
Okay, but what does all that really mean, and how does it affect how we diagnose a circuit? Two of these points are very important, and will come back to visit us as we examine other types of circuits. These are prin-ciples 4 and 5, which are otherwise known as Kirchhoff’s Law of Currentand Kirchhoff’s Law of Voltage.
Why Kirchhoff? Gustav Robert Kirchhoff was a German physicist who experimented with circuits in 1857. These experiments enabled him to establish his laws of current and volt-age.
Testing the Principles of Simple Circuits
Let’s start with the first principle, and work our way through each one, analyzing them as we go.
Principle 1: The circuit will use
all of the voltage available to it.That means, if the circuit has 12
volts applied, it will use all of that volt-age up by the time it gets to the ground, or negative side of the circuit. So what it boils down to is that the ground side of the circuit will have zero volts, because all 12 volts will have been used up by the circuit. Sound simple enough; let’s test it:
What we have here is a demonstra-tion board of a simple circuit (figure 1). It provides a power source (A), a load (B) and conductors (C). I’ve tapped into the conductors in a couple places to provide access points for testing (D).
According to principle 1, the circuit will use all of the voltage available to it. So if we connect our voltmeter negative lead to the negative battery terminal and probe the positive terminal, we’ll get the total applied voltage. In this case it’s 14.83 volts (figure 2).
When we move the positive lead to the negative terminal of the circuit, we get zero volts (figure 3). So the circuit is using all of the voltage applied;
by Steve Bodofsky
PUSHING ELECTRONS
Figure 1: This is a simple circuit, including a power source
(A), resistance (B), and conductors (C). And we’ve included
additional test points at a couple locations along the conduc-
tors (D).
Figure 2: The power source is applying 14.83 volts to the circuit.
GEARS September 2005 63
principle 1 appears to be correct.Principle 2: The voltage will be
the same throughout the power side of the circuit.
The power side of the circuit begins at the battery positive terminal, and
ends at the load. This prin-ciple says that the voltage should be the same at the battery as it is at the posi-tive feed at the load. Let’s try it:
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Figure 3: When we check the voltage after the load, we see that all of
the voltage is being used.
Figure 4: The voltage is equal everywhere on the positive side of the load.
When we checked the resistance of the
demonstration circuit, we had 5.5 ohms.
With 14.83 volts, Ohm’s Law says we should
have about 2.7 amps in the circuit (14.83 ÷
5.5 = 2.696…).
But when we checked the current flow,
there was only 228 milliamps; less than
1/10th of what Ohm’s Law tells us to
expect.
What happened? Is Ohm’s Law wrong?
No, Ohm’s Law is still correct. What hap-
pened is heat: When the bulb lit, it became
hot. This heat caused the filament in the
bulb to expand, and increased the separa-
tion between the atoms. This increased
separation, in turn, increased the resistance
in the bulb.
So, does that mean Ohm’s Law is use-
less? Hardly. Ohm’s Law is still a valuable
part of electrical diagnosis and circuit veri-
fication.
What it means is something that you’ve
heard many times before, right here in the
pages of GEARS: Always check electrical
circuits live. Use amperage or voltage drop
to measure resistance. A basic resistance
test won’t account for the changes that take
place when a circuit is loaded, such as heat.
Even a badly frayed wire may check out just
fine during a resistance check, but will fail a
voltage drop test or current draw test.
Resistance checks are great for verifying
that you have the right part or for a quick
check, but to test a circuit accurately, always
check circuits live, using a voltage drop test
or an amperage draw test.
But What Happened to Ohm’s Law?
64 GEARS September 2005
The voltage at the positive battery terminal is 14.83 volts. And, no matter where we move the positive lead — to the access point or the light receptacle — we still get 14.83 volts (figure 4). So voltage is indeed the same throughout the power side of the circuit.
Principle 3: The voltage will be the same throughout the ground side of the circuit.
The ground side of the circuit begins at the load, and ends at the bat-tery ground terminal. This principle says that the voltage should be the same everywhere along that part of the circuit; that is, it should be zero volts from the load to the battery negative terminal. Let’s try it:
The voltage at the negative terminal is zero volts. And, no matter where we move the positive meter lead, anywhere along the negative side of the circuit, the voltage is always zero (figure 5).
Principle 4: Amperage will be the same throughout the circuit.
This is Kirchhoff’s Law of Current,
which states that the total current that goes into a junction is equal to the total current that comes out of a junction. There is no extra current, and none is lost. Let’s try it:
To measure the current flow directly on the positive side of the circuit, we’ll disconnect the conductor from the posi-tive battery terminal, and connect the meter in series with the circuit (figure 6). According to the meter, this circuit has 228 milliamps flowing through it.
Now we’ll reconnect the conductor to the positive lead, and connect the meter in series at the negative lead (fig-ure 7). Still 228 milliamps. So current flow is the same wherever we measure it. Kirchhoff was right on target with his current law.
This is an important principle; what it means to you is that the current level in a circuit will be the same anywhere you check it: positive or ground side; close to the battery or near the load. Which means you can check current flow in a circuit pretty much anywhere
you can reach it.For example, when checking a trans-
mission solenoid, you can measure the current on the positive or negative side of the circuit, whichever is convenient. In some cases, only one side of the cir-cuit is available, such as a solenoid that grounds to the case. But since amper-age is the same throughout the circuit, you can measure it on either side; in this case, the positive side.
Principle 5: The total resistance of the circuit will be equal to the sum of each component’s resistance.
This is an easy one to check: First we’ll disconnect the circuit from the battery, and measure the resistance of the bulb, right at the receptacle (figure 8). 5.5 ohms.
Next, we’ll reconnect the recep-tacle, and measure the resistance at the conductors where they connect to the battery terminals (figure 9). Still 5.5 ohms. Okay, that works… but wait, there’s more…
One of the things we know about
Simple Circuits, and Kirchoff ’s Laws of Current and Voltage
Figure 5: Voltage is zero everywhere on the negative side
of the load.
Figure 6: The current flow on the positive side of the circuit is
228 milliamps.
Figure 8: The resistance of the light bulb is 5.5 ohms.Figure 7: The current flow on the negative side of the circuit is
also 228 milliamps, because current flow is equal everywhere
along the circuit.
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66 GEARS September 2005
electrical circuits is that resistance cre-ates a voltage drop. And when there’s more than one resis-tance in the circuit, the voltage drop will increase proportion-ally with resistance. And, when added together, the sum of the voltage drops around the circuit will be equal to the voltage drop for the entire circuit. This is Kirchhoff ’s Law of Voltage.
What that means is, if you start at a particu-lar point in a closed cir-cuit and go around the circuit, adding all the individual voltage drops until you reach the start-ing point, there would be no extra voltage, and all of the voltage would be accounted for.
Of course, this is a simple circuit, and we already proved that the resistance in the circuit will use all of the voltage (see principle 1).
But what if we had additional voltage left after the resistance? What would that mean? It would mean there’s unexpected resistance somewhere along the negative side of the circuit. It could indicate a bad or
dirty connection, frayed wires, or an interruption of any kind in the circuit.
If the resistance were in the positive side of the circuit, the voltage applied
to the load would be lower than the total circuit volt-age.
Ki rchhoff ’s Law of Voltage is the basis for a typical voltage drop test. Any voltage lost some-where other than a known resis-tance indicates an unexpected or unwanted resis-tance in the cir-cuit. This is why
checking circuit voltage drop is such a valuable test for verifying the condition of the circuit.
IMPORTANT! — When checking voltage drops in circuits, make sure you have the maximum current flowing that the circuit will normally be subjected to. If you test the circuit with less than the current it was designed to handle, the voltage drop may appear accept-able, even if there’s a problem with the circuit.
That’s enough for this issue. Next time we’ll move on to a slightly more complex circuit: the series circuit. Until then, keep on pushing those electrons!
Figure 9: The resistance of the entire circuit is also 5.5 ohms,
because the sum of the individual resistances will equal the
total circuit resistance.
1. All of these items are neces-
sary for a circuit, except:
A. Switch
B. Power Source
C. Load
D. Conductors
2. According to Kirchhoff’s Law
of Current:
A. Current multiplied by resistance
equals voltage.
B. It doesn’t matter where you mea-
sure current flow in a circuit.
C. As resistance goes up, current
goes down.
D. All of the above.
3. Kirchhoff’s Law of Voltage is
the basis for which type of test?
A. Amperage draw test
B. Voltage drop test
C. Mainline pressure test
D. Converter stall test
4. Tech A says leftover voltage
on the negative side of a circuit
indicates unwanted resistance in
the circuit.
Tech B says a voltage drop on the
positive side of a circuit indicates
unwanted resistance in the circuit.
Who’s right?
A. A only
B. B only
C. Both A and B
D. Neither A nor B
5. Tech A says a voltage drop test
is more accurate than a resistance
check, because it tests the circuit
under actual operating conditions.
Tech B says you should load the
circuit to its maximum operating
current during a voltage drop test.
Who’s right?
A. A only
B. B only
C. Both A and B
D. Neither A nor B
6. The relationship of voltage,
resistance and amperage shown in
the article demonstrates:
A. Ohm’s Law is wrong.
B. Resistance measurement isn’t an
accurate test.
C. The best way to test a circuit is
under normal operating conditions.
D. All of the above.
7. Tech A says you should always
check amperage draw on the posi-
tive side of the circuit.
Tech B says if you measure the
applied voltage and circuit resis-
tance, an amperage draw test is
unnecessary.
Who’s right?
A. A only
B. B only
C. Both A and B
D. Neither A nor B
8. A small amount of voltage on
the negative side of the circuit indi-
cates:
A. A normal condition.
B. The load resistance is too low, so
it’s letting some voltage through.
C. The applied voltage is too high,
so some voltage is pushing through.
D. There’s excess resistance in the
negative side of the circuit.
9. Tech A says, to measure volt-
age drop, you must connect your
meter in series with the circuit.
Tech B says, to measure current
draw directly through your meter,
you must connect your meter in
series with the circuit.
Who’s right?
A. A only
B. B only
C. Both A and B
D. Neither A nor B
10. The sum of the individual resis-
tances in a circuit will equal:
A. the total circuit resistance.
B. the total circuit voltage.
C. the total circuit amperage.
D. All of the above.
Test
Answer Key: 1.A, 2.B, 3.B, 4.C, 5.C, 6.C, 7.D, 8.D., 9.B, 10.A
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