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.

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 six-volt battery and a 12-ohm resistor, you need only apply Ohm’s Law and the formula for power.

Since I = V/R, the current is:

The power is:

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:

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:

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.

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.

Answering this question is just a matter of plugging numbers into the formula for resistors in parallel.

So = 4 .

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.

Recalling that P = I²R, 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 + = .

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.

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:

Now that you’ve looked at this two-step technique for dealing with circuits in parallel and in series, you should have no problem answering a range of other questions.

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:

Now that we know the current across each resistor, calculating the power dissipated is a straightforward application of the formula P = I²R:

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.

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.

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:

We can conclude that = 4 .

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.

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.

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 half-ampere 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.

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 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:

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 addresses the voltage drop of any closed loop in the circuit. It states:

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:

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.