What is the current through, and voltage drop across, each resistor in the following circuit?
In this problem, you are asked to find current and voltage drop in an electric circuit. Any time that you are given an electric circuit and asked for properties associated with that circuit (current, resistance, voltage, charge, capacitance) you will work the problem in one of two ways.
If the circuit is straightforward and you can track the flow of current, the best approach is to use a basic understanding of circuits to work the problem. If, however, the circuit contains a Wheatstone Bridge or is, in general, too complicated to track the current, then you will need to use Kirchhoff's Laws.
In this example, I will use a basic understanding of circuits to work the problem.
If you are asked for charge and capacitance, you work the circuit in exactly the same way as for current and resistance. The relation ΔV = IR that you use for resistors is replaced with ΔV = Q/C. Capacitors in parallel add according to C_{eq} = Σ_{i}C_{i}, while capacitors in series add according to 1/C_{eq} = Σ_{i}(1/C_{i}). The approach that you use, however, is identical.
Although Kirchhoff's Laws will work in any circuit, they have two disadvantages. First, they require a lot of linked algebraic equations. (In this problem you would need to jointly solve fourteen equations in fourteen unknowns.) Second, you do not need to know or learn anything at all about the physics of circuits to use Kirchhoff's Laws. So working circuits that way will strengthen your algebraic skills, but will not help you with conceptual questions about circuits. If you use the basic ideas behind circuits on practice problems, you will find that you can answer many mathematical and conceptual questions without having to fully go through all mathematical steps.
Although Kirchhoff's Laws will work in any circuit, they have two disadvantages. First, they require a lot of linked algebraic equations. (In this problem you would need to jointly solve fourteen equations in fourteen unknowns.) Second, you do not need to know or learn anything at all about the physics of circuits to use Kirchhoff's Laws. So working circuits that way will strengthen your algebraic skills, but will not help you with conceptual questions about circuits. If you use the basic ideas behind circuits on practice problems, you will find that you can answer many mathematical and conceptual questions without having to fully go through all mathematical steps.
The first thing you need to do in a circuit problem is reduce the circuit to a single equivalent resistance (or capacitance.) In other words, once you know the effective resistance of the circuit as a whole, you can calculate how much current the battery generates. So you must begin by simplifying the circuit.
Hint: The most frequent mistakes made in these problems come about from skipping this step or doing too many things at once. If you carefully reduce the circuit and record your drawings along the way (I like to put each drawing in a column along the left side of my paper so that I can work my way back up), you may be able to take shortcuts in later steps. But don't take shortcuts now!
It is handy for these problems to use colors to track sections of the circuit where current is the same. So before I begin to reduce the circuit, I use colored pencils or highlighters to trace the currents. A single color means that current is the same throughout that path.
Once the parallel portions of the circuit have been reduced to a single resistor, you can see that the net resistance of the circuit is equal to that of a 3.0, a 1.0 and a 2.0 Ω resistor in series with each other. There will be the same amount of current through the battery if these three resistors are replaced by a single 6.0 Ω resistor.
Make sure not to start in small pieces. I began with the only two elements of the circuit that are only in parallel or only in series. Any other portions of the circuit include both series and parallel components and so cannot be treated in a single step.
No. If you calculated a value of 1.0 Ω, you treated that portion of the circuit as if it were three resistors all in parallel. But in this case, you have two resistors in series with each other, and that series leg is in parallel with a third resistor. As suggested above, make sure to take small steps when you simplify the circuit.
No. If you calculated a value of 9.0 Ω, you treated that portion of the circuit as if it were three resistors all in series. But in this case, you have two resistors in series with each other, and that series leg is in parallel with a third resistor. As suggested above, make sure to take small steps when you simplify the circuit.
Yes. If you recognize that the right leg of the circuit is three resistors all in parallel, absolutely treat it in one step. In that case, you would replace that leg with a single 1.0 Ω resistor in the second drawing. If you do not see that it is three resistors all in parallel, it is fine to simplify it in two smaller steps as I demonstrate here.
The battery has no knowledge of the details of the circuit. Any circuit that has the same equivalent resistance will yield the same current through a given battery. In other words, if you take out this circuit and replace it with a single resistance of the "equivalent" amount, the battery will not know the difference and you will have the same amount of current. That single equivalent resistance is the starting place for solving a circuit problem.
Charge is conserved--in other words, current is the same in any series portion of the circuit (where there are no alternate paths.) It is true that electric potential energy is converted into other forms when current goes through a resistor, but the electrons themselves do not get used up. Current stays the same. At a junction, some of the current will follow each path but it still must add to the total value.
You will save steps if you begin to reduce the circuit by starting within parallel branches and working out. In this case, two 3.0 Ω resistors are in series with each other within a larger parallel section of the circuit. I cannot reduce that parallel section until I first reduce this series component within it. From the perspective of the battery and the current through each wire, it is equivalent to replace the two 3.0 Ω resistors with a single 6.0 Ω resistor.
In series,
R_{eq} = Σ_{i}R_{i} = 3.0 Ω + 3.0 Ω = 6.0 Ω.
In parallel,
1/R
R_{eq} = 1/(0.67 Ω^{-1}) = 1.5 Ω
In parallel,
1/R
R_{eq} = 1/(0.50 Ω^{-1}) = 2.0 Ω
In parallel,
1/R
R_{eq} = 1/(1.0 Ω^{-1}) = 1.0 Ω
In series,
R_{eq} = Σ_{i}R_{i} = 3.0 Ω + 1.0 Ω + 2.0 Ω= 6.0 Ω.
There are two ideas (charge is conserved, energy is conserved) and three equations that help you understand circuits.
As applied to circuits
Conservation of Charge gives: Current is the same through resistors in series.
Conservation of Energy gives: Voltage drop is the same across each leg of a parallel circuit.
The three equations are
R_{eq} = Σ_{i}R_{i}
1/R_{eq} = Σ_{i}1/R_{i}
ΔV = IR
You already used Conservation of Charge and the first two equations as you drew the pictures. To solve the problem, you will use Conservation of Energy, Conservation of Charge, and the third equation.
Some books use V instead of ΔV in this equation. I use ΔV to emphasize that this equation quantifies the change (or drop) in voltage across a resistor.
When the circuit reaches its steady state, the amount of energy (per charge) that the current gains as it goes across the battery is equal to the amount of energy (per charge) that it converts to other forms in the resistors. This balance is what determines how much is generated by the battery. This understanding can be restated as "voltage drop is the same across each leg of a parallel circuit." Regardless of which path the current takes, it must convert the same amount of energy (per charge) to other forms.
Once you have reduced a circuit to a single voltage source and an equivalent resistance, you can find the current through, and voltage drop across, each resistor by starting with the simplest (equivalent) drawing and working your way back up.
1.5 A of current is produced by the battery. The current everywhere that is color-coded green is also 1.5 A.
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Current is the same through resistors in series.
We now know the current and voltage drop for the 3.0 Ω resistor in the lower leg of the circuit. That resistor requires no further calculation.
Check: 4.5 V + 1.5 V + 3.0 V = 9.0 V. The voltage gain across the battery is balanced by conversions across the resistors.
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Voltage drop is the same across each leg of a parallel circuit.
We now know the current and voltage drop for the 3.0 Ω resistors in the blue and yellow branches of the circuit. No further calculation is required for those two resistors. Note the color coding of the currents and their values corresponds to the colors used to code the circuit. In other words, everywhere that the circuit is orange, the current has the value 0.50 A.
Check: 0.5 A + 1.0 A = 1.5 A. Current is conserved in and out of the junction for each of the parallel elements of the circuit.
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For the orange leg of the left parallel element, note that current is the same through resistors in series.
for each of the 3.0 O resistors in this branch. Check that 1.5 V + 1.5 V = 3.0 V, giving the correct total voltage drop across the orange leg.
For the purple leg of the right parallel element, note that voltage drop is the same across resistors in parallel.
for each of the 3.0 O resistors in this parallel element. Check: 0.5 A + 0.5 A = 1.0 A. Current is conserved in and out of the junction.
Current and voltage drop for each resistor in the circuit are now known. No further calculation is necessary.
As long as you work with one picture at a time, it is fine to reason out the currents and voltage drops. For example, in a parallel circuit with a 3.0 Ω resistor in one leg and a 6.0 Ω resistor in the other leg, you know that twice as much current will go through the leg with half the resistance. In other words, two thirds of the current will go through the 3.0 Ω resistor and one third of the current will go through the 6.0 Ω resistor.
The amount of current through the battery depends on the strength of the battery and the resistance of the circuit. Batteries do not put out a fixed amount of current. When you first connect a circuit, the current goes from zero to some steady state value. At that value, the amount of energy a charge gains from the battery is balanced by the amount of energy is loses through the resistors in its path. If the charge is given more energy than it loses, it accelerates and current grows. If it converts more energy than it gains, it gets slower and current decreases. When gains and losses balance, the current maintains its steady state value.
The battery increases voltage by 9.0 V. Therefore, there must also be a total voltage drop of 9.0 V in the circuit at steady state. Current gains and loses the same amount of energy over the circuit and therefore has a steady value.
The net or equivalent resistance of the circuit is 6.0 Ω. In other words, the current in steady state is determined by the 9.0 V of energy (per charge) from the battery and the 6.0 Ω resistance from the circuit.
There are no junctions in the green portion of the circuit. Everywhere color-coded green has a 1.5 A current.
Because there are no junctions in the green portion of the circuit, any current element that goes through the 3.0 Ω resistor must also go through the 1.0 Ω resistor and the 2.0 Ω resistor. In each case, that current is 1.5 A as found in the previous step.
The voltage drop across the 3.0 Ω resistor is 4.5 V.
The voltage drop across the 1.0 Ω resistor is 1.5 V.
The voltage drop across the 2.0 Ω resistor is 3.0 V.
As a current element follows a complete path around the circuit, it gains 9.0 V (Joules/Coulomb) across the battery. To be in steady state, it must also convert a total of 9.0 V to other forms of energy such as heat as it goes through the circuit elements.
The 2.0 Ω resistor of the previous picture is actually the equivalent resistance of the 6.0 and 3.0 Ω resistors in parallel with each other. Regardless of which of the paths a current element takes, it must convert a total of 3.0 V of energy per charge across that path in order for the circuit to be in steady state.
The 1.0 Ω resistor of the previous picture is actually the equivalent resistance of the 1.5 and 3.0 Ω resistors in parallel with each other. Regardless of which of the paths a current element takes, it must convert a total of 3.0 V of energy per charge across that path in order for the circuit to be in steady state.
We know that 1.5 A of current flows in the green portions of the circuit. When that current reaches the left parallel branch of the circuit, it divides. 0.50 A of current goes through the 6.0 Ω resistor, and 1.0 A of current goes through the 3.0 Ω resistor. There is more current in the branch with less resistance.
We know that 1.5 A of current flows in the green portions of the circuit. When that current reaches the right parallel branch of the circuit, it divides. 0.50 A of current goes through the 3.0 Ω resistor, and 1.0 A of current goes through the 1.5 Ω resistor. There is more current in the branch with less resistance.
Current does not get used up in resistors. Some of the energy associated with each electron is converted to other forms, but the electrons themselves remain and the current maintains a steady state value. Therefore, total current into a junction is equal to total current out of that junction.
The 6.0 Ω resistor in the previous step is actually the equivalent resistance of two 3.0 Ω resistors in series. Therefore, the current through the 3.0 Ω resistors is the same as the current through the 6.0 Ω resistor.
The 1.5 Ω resistor in the previous step is actually the equivalent resistance of two 3.0 Ω resistors in paralle. Therefore, the voltage drop across either of the 3.0 Ω resistors must be the same as the voltage drop across the 1.5 Ω resistor.
In this problem, you first use the understandings that current is the same through all series portions of a circuit and the equations for equivalent resistance to replace the given circuit with one that has the same effect from the perspective of the battery.
This break down of the circuit into one equivalent resistance is the key to working circuit problems successfully. Start within parallel components of the circuit and work out, and don't take more than one step at a time.
Once you know the equivalent effect of the circuit on the battery, you can use the understanding that voltage drop is the same across parallel legs of a circuit (conservation of energy) and the equation defining the voltage drop across a resistor to find the current and voltage drop for each resistor in parallel, and the understanding that current is the same through all series portions of a circuit to find the voltage drop across resistors in series.
Keeping track of your information is key. At each step, if you fill what you know into the appropriate intermediate drawing you are far less likely to make mistakes. If you replaced a series of resistors with a single resistor, the current through all is the same. Likewise, if you replaced a parallel portion of the circuit with a single resistor, the voltage drop across all is the same.
In this case, there is a current of 1.5 A through the battery. At each junction, the current splits. The amount of current through the branches is inversely proportional to the relative resistance of the branches. No current is used up, but electric potential energy is converted to other forms in each resistor.
In exactly the same way! The relation ΔV = IR that you use for resistors is replaced with ΔV = Q/C. Capacitors in parallel add according to C_{eq} = Σ_{i}C_{i}, while capacitors in series add according to 1/C_{eq} = Σ_{i}(1/C_{i}). The approach that you use, however, is identical.
Many complicated circuits can still be reduced using this approach. Look for pieces of the circuit that are clearly in parallel or clearly in series, and begin by simplifying those portions. In other words, reduce the circuit step by step. However, in some cases you may need to use Kirchhoff's Laws to solve for current and voltage drop. Click here to see an example problem using Kirchhoff's Laws.
If the lightbulbs in a circuit are identical, they all have the same resistance and the brightness of the bulbs is therefore an indicator of the current through the bulbs. Click here to see this circuit (of identical resistors) worked conceptually rather than mathematically.