


Circuits
Most SAT II Physics questions on circuits will show you
a circuit diagram and ask you questions about the current, resistance,
or voltage at different points in the circuit. These circuits will
usually consist of a power source and one or more resistors arranged
in parallel or in series. You will occasionally encounter other
circuit elements, such as a voltmeter, an ammeter, a fuse, or a
capacitor. Reading the diagrams is not difficult, but since there
will be a number of questions on the test that rely on diagrams,
it’s important that you master this skill. Here’s a very simple
circuit diagram:
Zigzags represent resistors, and a pair of parallel, unequal
lines represents a battery cell. The longer line is the positive
terminal and the shorter line is the negative terminal. That means
the current flows from the longer line around the circuit to the
shorter line. In the diagram above, the current flows counterclockwise.
Often, more than one set of unequal parallel lines are arranged
together; this just signifies a number of battery cells arranged
in series.
Example

You don’t really need to refer to the diagram in order
to solve this problem. As long as you know that there’s a circuit
with a sixvolt battery and a 12ohm resistor, you
need only apply Ohm’s Law and the formula for power.
Since I = V/R, the current
is:
The power is:
Resistors in Series
Two resistors are in series when they are
arranged one after another on the circuit, as in the diagram below.
The same amount of current flows first through one resistor and
then the other, since the current does not change over the length
of a circuit.
However, each resistor causes a voltage drop, and if there
is more than one resistor in the circuit, the sum of the voltage
drops across each resistor in the circuit is equal to the total voltage
drop in the circuit. The total resistance in a circuit with two
or more resistors in series is equal to the sum of the resistance
of all the resistors: a circuit would have the same resistance if
it had three resistors in series, or just one big resistor with
the resistance of the original three resistors put together. In
equation form, this principle is quite simple. In a circuit with
two resistors, and , in series, the total resistance, is:
Example

What is the current in the circuit?
We can determine the current in the circuit by applying Ohm’s
Law: I = V/R. We know what V is,
but we need to calculate the total resistance in the circuit by
adding together the individual resistances of the two resistors
in series:
When we know the total resistance in the circuit, we can
determine the current through the circuit with a simple application
of Ohm’s Law:
What is the voltage drop across each resistor?
Determining the voltage drop across an individual resistor
in a series of resistors simply requires a reapplication of Ohm’s
Law. We know the current through the circuit, and we know the resistance
of that individual resistor, so the voltage drop across that resistor
is simply the product of the current and the resistance. The voltage
drop across the two resistors is:
Note that the voltage drop across the two resistors is 10 V
+ 20 V = 30 V, which is the total voltage
drop across the circuit.
Resistors in Parallel
Two resistors are in parallel when the circuit
splits in two and one resistor is placed on each of the two branches.
In this circumstance, it is often useful to calculate
the equivalent resistance as if there were only one resistor, rather
than deal with each resistor individually. Calculating the equivalent
resistance of two or more resistors in parallel is a little more
complicated than calculating the total resistance of two or more
resistors in series. Given two resistors, and , in parallel, the equivalent
resistance, , is:
When a circuit splits in two, the current is divided between
the two branches, though the current through each resistor will
not necessarily be the same. The voltage drop must be the same across
both resistors, so the current will be stronger for a weaker resistor,
and vice versa.
Example

What is the total resistance in the circuit?
Answering this question is just a matter of plugging numbers
into the formula for resistors in parallel.
So = 4 .
What is the current running through R1 and R2?
We know that the total voltage drop is 12 V,
and since the voltage drop is the same across all the branches of
a set of resistors in parallel, we know that the voltage drop across
both resistors will be 12 V. That means we just need
to apply Ohm’s Law twice, once to each resistor:
If we apply Ohm’s Law to the total resistance in the system,
we find that = (12 V)/(4 ) = 3 A. As we might expect, the total current
through the system is the sum of the current through each of the
two branches. The current is split into two parts when it branches into
the resistors in parallel, but the total current remains the same
throughout the whole circuit. This fact is captured in the junction
rule we will examine when we look at Kirchhoff’s Rules.
What is the power dissipated in the resistors?
Recalling that P = I^{2}R,
we can solve for the power dissipated through each resistor individually,
and in the circuit as a whole. Let be the power dissipated in , the power dissipated
in , and the power dissipated
in .
Note that + = .
Circuits with Resistors in Parallel and in Series
Now that you know how to deal with resistors in parallel
and resistors in series, you have all the tools to approach a circuit
that has resistors both in parallel and in series. Let’s take a look
at an example of such a circuit, and follow two important steps
to determine the total resistance of the circuit.
 Determine the equivalent resistance of the resistors in parallel. We’ve already learned to make such calculations. This one is no different:
So the equivalent resistance is 6 . In effect, this means that the two resistors
in parallel have the same resistance as if there were a single 6 resistor in their place. We can redraw the
diagram to capture this equivalence:
 Treating the equivalent resistance of the resistors in parallel as a single resistor, calculate the total resistance by adding resistors in series. The diagram above gives us two resistors in series. Calculating the total resistance of the circuit couldn’t be easier:
Now that you’ve looked at this twostep technique for
dealing with circuits in parallel and in series, you should have
no problem answering a range of other questions.
Example

What is the current running through each resistor?
We know that resistors in series do not affect the current,
so the current through is the same as the total
current running through the circuit. Knowing the total resistance
of the circuit and the voltage drop through the circuit, we can
calculate the circuit’s total current by means of Ohm’s Law:
Therefore, the current through is 3 A.
But be careful before you calculate the current through and : the voltage drop across
these resistors is not the total voltage drop of 30 V.
The sum of the voltage drops across and the two resistors in parallel is 30 V,
so the voltage drop across just the resistors in parallel is less
than 30 V.
If we treat the resistors in parallel as a single equivalent
resistor of 6 , we can calculate the
voltage drop across the resistors by means of Ohm’s Law:
Now, recalling that current is divided unevenly between
the branches of a set of resistors in parallel, we can calculate
the current through and in the familiar way:
What is the power dissipated across each resistor?
Now that we know the current across each resistor, calculating
the power dissipated is a straightforward application of the formula P
= I^{2}R:
Common Devices in Circuits
In real life (and on SAT II Physics) it is
possible to hook devices up to a circuit that will read off the
potential difference or current at a certain point in the circuit.
These devices provide SAT II Physics with a handy means of testing
your knowledge of circuits.
Voltmeters and Ammeters
A voltmeter, designated:
measures the voltage across a wire. It is connected in
parallel with the stretch of wire whose voltage is being measured,
since an equal voltage crosses both branches of two wires connected
in parallel.
An ammeter, designated:
is connected in series. It measures the current passing
through that point on the circuit.
Example

What does the ammeter read?
Since the ammeter is not connected in parallel with any
other branch in the circuit, the reading on the ammeter will be
the total current in the circuit. We can use Ohm’s Law to determine
the total current in the circuit, but only if we first determine
the total resistance in the circuit.
This circuit consists of resistors in parallel and in
series, an arrangement we have looked at before. Following the same
two steps as we did last time, we can calculate the total resistance
in the circuit:
 Determine the equivalent resistance of the resistors in parallel.
We can conclude that = 4 .
 Treating the equivalent resistance of the resistors in parallel as a single resistor, calculate the total resistance by adding resistors in series.
Given that the total resistance is 9 and the total voltage is 9 V,
Ohm’s Law tells us that the total current is:
The ammeter will read 1 A.
What does the voltmeter read?
The voltmeter is connected in parallel with and , so it will register
the voltage drop across these two resistors. Recall that the voltage
drop across resistors in parallel is the same for each resistor.
We know that the total voltage drop across the circuit
is 9 V. Some of this voltage drop will take place across , and the rest of the voltage drop will
take place across the resistors in parallel. By calculating the
voltage drop across and subtracting from 9 V,
we will have the voltage drop across the resistors in parallel,
which is what the voltmeter measures.
If the voltage drop across is 5 V, then the voltage drop
across the resistors in parallel is 9 V – 5 V
= 4 V. This is what the voltmeter reads.
Fuses
A fuse burns out if the current in a circuit
is too large. This prevents the equipment connected to the circuit
from being damaged by the excess current. For example, if the ammeter
in the previous problem were replaced by a halfampere fuse, the
fuse would blow and the circuit would be interrupted.
Fuses rarely come up on SAT II Physics. If a question
involving fuses appears, it will probably ask you whether or not
the fuse in a given circuit will blow under certain circumstances.
Kirchhoff’s Rules
Gustav Robert Kirchhoff came up with two simple rules
that simplify many complicated circuit problems. The junction
rule helps us to calculate the current through resistors
in parallel and other points where a circuit breaks into several
branches, and the loop rule helps us to calculate the
voltage at any point in a circuit. Let’s study Kirchhoff’s Rules
in the context of the circuit represented below:
Before we can apply Kirchhoff’s Rules, we have to draw
arrows on the diagram to denote the direction in which we will follow
the current. You can draw these arrows in any direction
you please—they don’t have to denote the actual direction of the
current. As you’ll see, so long as we apply Kirchhoff’s Rules correctly,
it doesn’t matter in what directions the arrows point. Let’s draw
in arrows and label the six vertices of the circuit:
We repeat, these arrows do not point in the actual direction
of the current. For instance, we have drawn the current flowing
into the positive terminal and out of the negative terminal of , contrary to how we know the current must
flow.
The Junction Rule
The junction rule deals with “junctions,” where a circuit
splits into more than one branch, or when several branches reunite
to form a single wire. The rule states:
The current coming into a junction equals the
current coming out.
This rule comes from the conservation of charge: the charge
per unit time going into the junction must equal the charge per
unit time coming out. In other words, when a circuit separates into
more than one branch—as with resistors in parallel—then the total
current is split between the different branches.
The junction rule tells us how to deal with resistors
in series and other cases of circuits branching in two or more directions.
If we encounter three resistors in series, we know that the sum
of the current through all three resistors is equal to the current
in the wire before it divides into three parallel branches.
Let’s apply the junction rule to the junction at B in
the diagram we looked at earlier.
According to the arrows we’ve drawn, the current in the
diagram flows from A into B across and flows out of B in
two branches: one across toward E and
the other toward C. According to the
junction rule, the current flowing into B must
equal the current flowing out of B.
If we label the current going into B as and the current going out of B toward E as , we can conclude that the current going
out of B toward C is – . That way, the current
flowing into B is and the current flowing out of B is + ( – ) = .
The Loop Rule
The loop rule addresses the voltage drop of any closed
loop in the circuit. It states:
The sum of the voltage drops around a closed
loop is zero.
This is actually a statement of conservation of energy:
every increase in potential energy, such as from a battery, must
be balanced by a decrease, such as across a resistor. In other words,
the voltage drop across all the resistors in a closed loop is equal
to the voltage of the batteries in that loop.
In a normal circuit, we know that when the current crosses
a resistor, R, the voltage drops by IR,
and when the current crosses a battery, V,
the voltage rises by V. When we trace
a loop—we can choose to do so in the clockwise direction or the
counterclockwise direction—we may sometimes find ourselves tracing
the loop against the direction of the arrows we drew. If we cross
a resistor against the direction of the arrows, the voltage rises by IR.
Further, if our loop crosses a battery in the wrong direction—entering
in the positive terminal and coming out the negative terminal—the
voltage drops by V. To summarize:
 Voltage drops by IR when the loop crosses a resistor in the direction of the current arrows.
 Voltage rises by IR when the loop crosses a resistor against the direction of the current arrows.
 Voltage rises by V when the loop crosses a battery from the negative terminal to the positive terminal.
 Voltage drops by V when the loop crosses a battery from the positive terminal to the negative terminal.
Let’s now put the loop rule to work in sorting out the
current that passes through each of the three resistors in the diagram
we looked at earlier. When we looked at the junction rule, we found
that we could express the current from A to B—and
hence the current from E to D to A—as , the current from B to E as , and the current from B to C—and
hence the current from C to F to E—as – . We have two variables
for describing the current, so we need two equations in order to
solve for these variables. By applying the loop rule to two different
loops in the circuit, we should be able to come up with two different
equations that include the variables we’re looking for. Let’s begin
by examining the loop described by ABED.
Remember that we’ve labeled the current between A and B as and the current between B and E as . Because the current flowing from E to A is
the same as that flowing from A to B,
we know this part of the circuit also has a current of .
Tracing the loop clockwise from A,
the current first crosses and the voltage drops by . Next it crosses and the voltage drops by . Then the current crosses , and the voltage rises by 12 V.
The loop rule tells us that the net change in voltage is zero across
the loop. We can express these changes in voltage as an equation,
and then substitute in the values we know for , , and :
Now let’s apply the loop rule to the loop described by BCFE.
Tracing the loop clockwise from B,
the arrows cross , but in the wrong direction,
from positive to negative, meaning that the voltage drops by 8 V.
Next, the current crosses , with an additional voltage
drop of . Finally, it crosses , but in the opposite direction of the arrows,
so the current goes up by . Now we can construct
a second equation:
Plugging this solution for into the earlier equation of 4 + 3 = 12, we get:
So the current across is 28/13 A.
With that in mind, we can determine the current across and by plugging the value
for into the equations we derived earlier:
The negative value for the current across means that the current actually flows in
the opposite direction of the arrow we drew. This makes perfect
sense when we consider that current should normally flow out of
the positive terminal and into the negative terminal of battery .
It doesn’t matter how you draw the current arrows on the
diagram, because if you apply Kirchhoff’s Rules correctly, you will
come up with negative values for current wherever your current arrows
point in the opposite direction of the true current. Once you have
done all the math in accordance with Kirchhoff’s Rules, you will
quickly be able to determine the true direction of the current.
