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Electric Circuits: Parallel and series networks (Grade 11)

Module by: Free High School Science Texts Project. E-mail the author

Series and parallel networks of resistors

Now that you know how to handle simple series and parallel circuits, you are ready to tackle problems like this:

Figure 1: An example of a series-parallel network. The dashed boxes indicate parallel sections of the circuit.
Figure 1 (PG11C9_016.png)

It is relatively easy to work out these kind of circuits because you use everything you have already learnt about series and parallel circuits. The only difference is that you do it in stages. In Figure 1, the circuit consists of 2 parallel portions that are then in series with 1 resistor. So, in order to work out the equivalent resistance, you start by calculating the total resistance of the parallel portions and then add up all the resistances in series. If all the resistors in Figure 1 had resistances of 10 ΩΩ, we can calculate the equivalent resistance of the entire circuit.

We start by calculating the total resistance of Parallel Circuit 1.

Figure 2
Figure 2 (PG11C9_017.png)

The value of Rp1Rp1 is:

1 R p 1 = 1 R 1 + 1 R 2 + 1 R 3 R p 1 = 1 10 + 1 10 + 1 10 - 1 = 1 + 1 + 1 10 - 1 = 3 10 - 1 = 3 , 33 Ω 1 R p 1 = 1 R 1 + 1 R 2 + 1 R 3 R p 1 = 1 10 + 1 10 + 1 10 - 1 = 1 + 1 + 1 10 - 1 = 3 10 - 1 = 3 , 33 Ω
(1)

We can similarly calculate the total resistance of Parallel Circuit 2:

1 R p 2 = 1 R 5 + 1 R 6 + 1 R 7 R p 2 = 1 10 + 1 10 + 1 10 - 1 = 1 + 1 + 1 10 - 1 = 3 10 - 1 = 3 , 33 Ω 1 R p 2 = 1 R 5 + 1 R 6 + 1 R 7 R p 2 = 1 10 + 1 10 + 1 10 - 1 = 1 + 1 + 1 10 - 1 = 3 10 - 1 = 3 , 33 Ω
(2)

Figure 3
Figure 3 (PG11C9_018.png)

This has now being simplified to a simple series circuit and the equivalent resistance is:

R = R p 1 + R 4 + R p 2 = 10 3 + 10 + 10 3 = 100 + 30 + 100 30 = 230 30 = 7 , 66 Ω R = R p 1 + R 4 + R p 2 = 10 3 + 10 + 10 3 = 100 + 30 + 100 30 = 230 30 = 7 , 66 Ω
(3)

The equivalent resistance of the circuit in Figure 1 is 7,66ΩΩ.

Series and parallel networks

Determine the equivalent resistance of the following circuits:

  1. Figure 4
    Figure 4 (PG11C9_019.png)
  2. Figure 5
    Figure 5 (PG11C9_020.png)
  3. Figure 6
    Figure 6 (PG11C9_021.png)

Wheatstone bridge

Another method of finding an unknown resistance is to use a Wheatstone bridge. A Wheatstone bridge is a measuring instrument that is used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. Its operation is similar to the original potentiometer except that in potentiometer circuits the meter used is a sensitive galvanometer.

Note: Interesting Fact :

The Wheatstone bridge was invented by Samuel Hunter Christie in 1833 and improved and popularized by Sir Charles Wheatstone in 1843.

Figure 7
Figure 7 (PG11C9_022.png)

In the circuit of the Wheatstone bridge, RxRx is the unknown resistance. R1R1, R2R2 and R3R3 are resistors of known resistance and the resistance of R2R2 is adjustable. If the ratio of R2R2:R1R1 is equal to the ratio of RxRx:R3R3, then the voltage between the two midpoints will be zero and no current will flow between the midpoints. In order to determine the unknown resistance, R2R2 is varied until this condition is reached. That is when the voltmeter reads 0 V.

Exercise 1: Wheatstone bridge

What is the resistance of the unknown resistor RxRx in the diagram below if R1R1=4ΩΩ  R2R2=8ΩΩ and R3R3=6ΩΩ.

Figure 8
Figure 8 (PG11C9_023.png)

Solution

  1. Step 1. Determine how to approach the problem :

    The arrangement is a Wheatstone bridge. So we use the equation:

    R x : R 3 = R 2 : R 1 R x : R 3 = R 2 : R 1
    (4)
  2. Step 2. Solve the problem :
    R x : R 3 = R 2 : R 1 R x : 6 = 8 : 4 R x = 12 Ω R x : R 3 = R 2 : R 1 R x : 6 = 8 : 4 R x = 12 Ω
    (5)
  3. Step 3. Write the final answer :

    The resistance of the unknown resistor is 12 ΩΩ.

Power in electric circuits

In addition to voltage and current, there is another measure of free electron activity in a circuit: power. Power is a measure of how rapidly a standard amount of work is done. In electric circuits, power is a function of both voltage and current:

Definition 1: Electrical Power

Electrical power is calculated as:

P = I · V P = I · V
(6)

Power (PP) is exactly equal to current (II) multiplied by voltage (VV) and there is no extra constant of proportionality. The unit of measurement for power is the Watt (abbreviated W).

Note: Interesting Fact :

It was James Prescott Joule, not Georg Simon Ohm, who first discovered the mathematical relationship between power dissipation and current through a resistance. This discovery, published in 1841, followed the form of the equation:
P = I 2 R P = I 2 R
(7)
and is properly known as Joule's Law. However, these power equations are so commonly associated with the Ohm's Law equations relating voltage, current, and resistance that they are frequently credited to Ohm.

Investigation : Equivalence

Use Ohm's Law to show that:

P = V I P = V I
(8)

is identical to

P = I 2 R P = I 2 R
(9)

and

P = V 2 R P = V 2 R
(10)

Summary

  1. Ohm's Law states that the amount of current through a conductor, at constant temperature, is proportional to the voltage across the resistor. Mathematically we write V=I·RV=I·R
  2. Conductors that obey Ohm's Law are called ohmic conductors; those that do not are called non-ohmic conductors.
  3. We use Ohm's Law to calculate the resistance of a resistor. R=VIR=VI
  4. The equivalent resistance of resistors in series (RsRs) can be calculated as follows: Rs=R1+R2+R3+...+RnRs=R1+R2+R3+...+Rn
  5. The equivalent resistance of resistors in parallel (RpRp) can be calculated as follows: 1Rp=1R1+1R2+1R3+...+1Rn1Rp=1R1+1R2+1R3+...+1Rn
  6. Real batteries have an internal resistance.
  7. Wheatstone bridges can be used to accurately determine the resistance of an unknown resistor.

End of chapter exercise

  1. Calculate the current in the following circuit and then use the current to calculate the voltage drops across each resistor.
    Figure 9
    Figure 9 (PG11C9_024.png)
  2. Explain why a voltmeter is placed in parallel with a resistor.
  3. Explain why an ammeter is placed in series with a resistor.
  4. [IEB 2001/11 HG1] - Emf
    1. Explain the meaning of each of these two statements:
      1. “The current through the battery is 50 mA.”
      2. “The emf of the battery is 6 V.”
    2. A battery tester measures the current supplied when the battery is connected to a resistor of 100 ΩΩ. If the current is less than 50 mA, the battery is “flat” (it needs to be replaced). Calculate the maximum internal resistance of a 6 V battery that will pass the test.
  5. [IEB 2005/11 HG] The electric circuit of a torch consists of a cell, a switch and a small light bulb, as shown in the diagram below.
    Figure 10
    Figure 10 (PG11C9_025.png)
    The electric torch is designed to use a D-type cell, but the only cell that is available for use is an AA-type cell. The specifications of these two types of cells are shown in the table below:
    Table 1
    CellemfAppliance for which it is designedCurrent drawn from cell when connected to the appliance for which it is designed
    D1,5 Vtorch300 mA
    AA1,5 VTV remote control30 mA
    What is likely to happen and why does it happen when the AA-type cell replaces the D-type cell in the electric torch circuit?
    Table 2
     What happensWhy it happens
    (a)the bulb is dimmerthe AA-type cell has greater internal resistance
    (b)the bulb is dimmerthe AA-type cell has less internal resistance
    (c)the brightness of the bulb is the samethe AA-type cell has the same internal resistance
    (d)the bulb is brighterthe AA-type cell has less internal resistance
  6. [IEB 2005/11 HG1] A battery of emf εε and internal resistance r = 25 ΩΩ is connected to this arrangement of resistors.
    Figure 11
    Figure 11 (PG11C9_026.png)
    The resistances of voltmeters V11 and V22 are so high that they do not affect the current in the circuit.
    1. Explain what is meant by “the emf of a battery”. The power dissipated in the 100 ΩΩ resistor is 0,81 W.
    2. Calculate the current in the 100 ΩΩ resistor.
    3. Calculate the reading on voltmeter V22.
    4. Calculate the reading on voltmeter V11.
    5. Calculate the emf of the battery.
  7. [SC 2003/11] A kettle is marked 240 V; 1 500 W.
    1. Calculate the resistance of the kettle when operating according to the above specifications.
    2. If the kettle takes 3 minutes to boil some water, calculate the amount of electrical energy transferred to the kettle.
  8. [IEB 2001/11 HG1] - Electric Eels Electric eels have a series of cells from head to tail. When the cells are activated by a nerve impulse, a potential difference is created from head to tail. A healthy electric eel can produce a potential difference of 600 V.
    1. What is meant by “a potential difference of 600 V”?
    2. How much energy is transferred when an electron is moved through a potential difference of 600 V?

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