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Ohm’s Law and the Combination of Resistors

It turns out that the current through most resistors is approximately proportional to the electric potential across them.
Ohm's Law
If / is the current through a given resistor and A V is the potential across the resistor, then
where R is the resistance of the resistor, a measure of how difficult it is for charge to (low through it. The unit for R is


In the next example we encounter a combination of resistors. Several resistors in a circuit are often either in series or in parallel. Several resistors are in series if a charge coming from the source must go through each of them before going back to the source. Several resistors are in parallel if a charge may go through any of them before going back to the source.


Two resistors (R1 = 10 Ω and R2 = 20 Ω) are connected in series with a potential source (9 volts). What is the current through resistor 1?



First, let’s DRAW A DIAGRAM of the circuit (Figure 15-7).
We label the lower wire 0 volts. There is 9-volt jump across the voltage source, so the upper wire is labeled 9 volts. We label the wire between the two resistors with the potential Vm. The current is the same through both resistors and the source.
The equivalent water course is shown in Figure 15-8 in which the current in the top trough is the same as the current through both waterfalls and through the bottom trough.

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Figure 15-7 Figure 15-8


If several resistors are in series, then the same current flow through all of them.


Do you see why?
Applying Ohm’s law to the first resistor gives

9VVm = IR1


and applying Ohm’s law to the second resistor gives

Vm – 0 = IR2


Substituting the expression for Vm gives

9VIR2 = IR1


9V = I(R1 + R2)


I = 9V/(R1 + R2)


= 9V/(10 Ω + 20 Ω)


= 0.3A


In the circuit above, what is the potential drop across resistor 1?



If we apply Ohm’s law to resistor 1, then we have


ΔV1 = I1R1
= (0.3A)(10 Ω)
= 3V

Generally, we do not have to go through as much trouble as we did in Example 1 because there are two rules for combining resistors:

If several resistors (/?,. /?,, and so on) arc in series, then we can replace them with one resistor whose resistance is the sum


If several resistors (R1R2… and so on) are in parallel then we can replace them with one resistor with resistance Rt. where




In the circuit shown (Figure 15-9), we have R1 = 20 Ω, R2 = 30 Ω, and R3 = 60 Ω, and the potential source is 12 volts.

  1. What is the current through resistor R2?
  2. What is the current through the potential source?

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Figure 15-9

  1. First, let’s label the negative terminal of the source 0 V and the positive terminal 12 V. The potential all along the left wire is 0 volts, and along the right wire is 12 volts (Figure 15-10).

    This gives us the potential drop across R2:
ΔV2 = 12V
I2 = ΔV2/R2
= 12V/30Ω
= 0.4A


  1. Note that the current splits into three parts through the resistors. To obtain the total current, we need to combine resistors to obtain an equivalent circuit (Figure 15-11):


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Figure 15-10 Figure 15-11


Figure 15-12 shows the analogous water-course.

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Figure 15-12



If several resistors are in parallel then the same potential exists across all of them.


When faced with a question concerning a circuit, there is no general procedure, which always leads to an answer, but here are some things to try:
  1. Label the lower end of the battery 0 V.
  2. Label other wires with voltages.
  3. Label the currents going through the wires.
  4. Combine resistors.
  5. Apply Ohm’s law.


Consider the circuit shown in Figure 15-13, where R1 = 1 Ω, R2 = R3 = 4 Ω, and Vcell = 6 V.

  1. What is the current going through resistor 1?
  2. How does I1 change if points A and B are connected with a wire?

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Figure 15-13

  1. First we draw in the current and voltages (Figure 15-14), but this does not seem to get anywhere, because we do not know a potential difference across resistor 1 to apply Ohm’s law. (Many students make the mistake here of using 6 volts and 1 Ω in Ohm’s law, but the potential difference across resistor 1 is not 6 volts.)

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Figure 15-14


Figure 15-15 shows the result of combining the two parallel resistors. Figure 15-16 shows the final equivalent circuit, by combining the two resistors in Figure 15-15. The total current is (by Ohm’s law, finally) 2 A. The total current is the same as the current through resistor 1 (see Figure 15-14), so the answer to part a is 2 A.


..\art 15 jpg\figure 15-to.jpg ..\art 15 jpg\figure 15-tp.jpg
Figure 15-15 Figure 15-16
  1. If we connect points A and B by a wire, then they have the same potential, and this is shown in Figure 15-17. No current flows through resistors 2 and 3, and resistor 1 has the full 6 V across it. The current through it (by Ohm’s law) is 6 A.

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Figure 15-17

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