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Electric Circuits - Grade 11

Introduction

The study of electrical circuits is essential to understand the technology that uses electricity in the real-world. This includes electricity being used for the operation of electronic devices like computers.

Ohm's Law

Definition of Ohm's Law

Experiment : Ohm's Law

Aim:

In this experiment we will look at the relationship between the current going through a resistor and the potential difference (voltage) across the same resistor.

Figure 1
Figure 1 (PG11C9_001.png)

Method:

  1. Set up the circuit according to the circuit diagram, starting with just one cell.
  2. Draw the following table in your lab book.
    Table 1
    Voltage, VV (V)Current, II (A)
    1,5 
    3,0 
    4,5 
    6,0 
  3. Get your teacher to check the circuit before turning the power on.
  4. Measure the current.
  5. Add one more 1,5 V cell to the circuit and measure the current again.
  6. Repeat until you have four cells and you have completed your table.
  7. Draw a graph of voltage versus current.

Results:

  1. Does your experimental results verify Ohm's Law? Explain.
  2. How would you go about finding the resistance of an unknown resistor using only a power supply, a voltmeter and a known resistor R0R0?

Activity : Ohm's Law

If you do not have access to the equipment necessary for the Ohm's Law experiment, you can do this activity.

Table 2
Voltage, VV (V) Current, II (A)
3,0 0,4
6,0 0,8
9,0 1,2
12,0 1,6
  1. Plot a graph of voltage (on the xx-axis) and current (on the yy-axis).

Conclusions:

  1. What type of graph do you obtain (straight line, parabola, other curve)
  2. Calculate the gradient of the graph.
  3. Do your experimental results verify Ohm's Law? Explain.
  4. How would you go about finding the resistance of an unknown resistor using only a power supply, a voltmeter and a known resistor R0R0?

An important relationship between the current, voltage and resistance in a circuit was discovered by Georg Simon Ohm and is called Ohm's Law.

Definition 1: Ohm's Law

The amount of electric current through a metal conductor, at a constant temperature, in a circuit is proportional to the voltage across the conductor. Mathematically, Ohm's Law is written:

V = R · I . V = R · I .
(1)

Ohm's Law tells us that if a conductor is at a constant temperature, the current flowing through the conductor is proportional to the voltage across it. This means that if we plot voltage on the xx-axis of a graph and current on the yy-axis of the graph, we will get a straight-line. The gradient of the straight-line graph is related to the resistance of the conductor.

Figure 2
Figure 2 (PG11C9_002.png)

Ohmic and non-ohmic conductors

As you have seen, there is a mention of constant temperature when we talk about Ohm's Law. This is because the resistance of some conductors changes as their temperature changes. These types of conductors are called non-ohmic conductors, because they do not obey Ohm's Law. As can be expected, the conductors that obey Ohm's Law are called ohmic conductors. A light bulb is a common example of a non-ohmic conductor. Nichrome wire is an ohmic conductor.

In a light bulb, the resistance of the filament wire will increase dramatically as it warms from room temperature to operating temperature. If we increase the supply voltage in a real lamp circuit, the resulting increase in current causes the filament to increase in temperature, which increases its resistance. This effectively limits the increase in current. In this case, voltage and current do not obey Ohm's Law.

The phenomenon of resistance changing with variations in temperature is one shared by almost all metals, of which most wires are made. For most applications, these changes in resistance are small enough to be ignored. In the application of metal lamp filaments, which increase a lot in temperature (up to about 1000C, and starting from room temperature) the change is quite large.

In general non-ohmic conductors have plots of voltage against current that are curved, indicating that the resistance is not constant over all values of voltage and current.

Figure 3
Figure 3 (PG11C9_003.png)

Experiment : Ohmic and non-ohmic conductors

Repeat the experiment as decribed in the previous section. In this case use a light bulb as a resistor. Compare your results to the ohmic resistor.

Using Ohm's Law

We are now ready to see how Ohm's Law is used to analyse circuits.

Consider the circuit with an ohmic resistor, RR. If the resistor has a resistance of 5 ΩΩ and voltage across the resistor is 5 V, then we can use Ohm's law to calculate the current flowing through the resistor.

Figure 4
Figure 4 (PG11C9_004.png)

Ohm's law is:

V = R · I V = R · I
(2)

which can be rearranged to:

I = V R I = V R
(3)

The current flowing through the resistor is:

I = V R = 5 V 5 Ω = 1 A I = V R = 5 V 5 Ω = 1 A
(4)

Exercise 1: Ohm's Law

Figure 5
Figure 5 (PG11C9_005.png)

The resistance of the above resistor is 10 ΩΩ and the current going through the resistor is 4 A. What is the potential difference (voltage) across the resistor?

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

    It is an Ohm's Law problem. So we use the equation:

    V = R · I V = R · I
    (5)
  2. Step 2. Solve the problem :
    V = R · I = ( 10 ) ( 4 ) = 40 V V = R · I = ( 10 ) ( 4 ) = 40 V
    (6)
  3. Step 3. Write the final answer :

    The voltage across the resistor is 40 V.

Ohm's Law

  1. Calculate the resistance of a resistor that has a potential difference of 8 V across it when a current of 2 A flows through it.
  2. What current will flow through a resistor of 6 ΩΩ when there is a potential difference of 18 V across its ends?
  3. What is the voltage across a 10 ΩΩ resistor when a current of 1,5 A flows though it?

Resistance

In Grade 10, you learnt about resistors and were introduced to circuits where resistors were connected in series and circuits where resistors were connected in parallel. In a series circuit there is one path for the current to flow through. In a parallel circuit there are multiple paths for the current to flow through.

Figure 6
Figure 6 (PG11C9_006.png)

Equivalent resistance

When there is more than one resistor in a circuit, we are usually able to calculate the total combined resitance of all the resistors. The resistance of the single resistor is known as equivalent resistance.

Equivalent Series Resistance

Consider a circuit consisting of three resistors and a single cell connected in series.

Figure 7
Figure 7 (PG11C9_007.png)

The first principle to understand about series circuits is that the amount of current is the same through any component in the circuit. This is because there is only one path for electrons to flow in a series circuit. From the way that the battery is connected, we can tell which direction the current will flow. We know that current flows from positive to negative, by convention. Current in this circuit will flow in a clockwise direction, from point A to B to C to D and back to A.

So, how do we use this knowledge to calculate the total resistance in the circuit?

We know that in a series circuit the current has to be the same in all components. So we can write:

I = I 1 = I 2 = I 3 I = I 1 = I 2 = I 3
(7)

We also know that total voltage of the circuit has to be equal to the sum of the voltages over all three resistors. So we can write:

V = V 1 + V 2 + V 3 V = V 1 + V 2 + V 3
(8)

Finally, we know that Ohm's Law has to apply for each resistor individually, which gives us:

V 1 = I 1 · R 1 V 2 = I 2 · R 2 V 3 = I 3 · R 3 V 1 = I 1 · R 1 V 2 = I 2 · R 2 V 3 = I 3 · R 3
(9)

Therefore:

V = I 1 · R 1 + I 2 · R 2 + I 3 · R 3 V = I 1 · R 1 + I 2 · R 2 + I 3 · R 3
(10)

However, because

I = I 1 = I 2 = I 3 I = I 1 = I 2 = I 3
(11)

, we can further simplify this to:

V = I · R 1 + I · R 2 + I · R 3 = I ( R 1 + R 2 + R 3 ) V = I · R 1 + I · R 2 + I · R 3 = I ( R 1 + R 2 + R 3 )
(12)

Further, we can write an Ohm's Law relation for the entire circuit:

V = I · R V = I · R
(13)

Therefore:

V = I ( R 1 + R 2 + R 3 ) I · R = I ( R 1 + R 2 + R 3 ) R = R 1 + R 2 + R 3 V = I ( R 1 + R 2 + R 3 ) I · R = I ( R 1 + R 2 + R 3 ) R = R 1 + R 2 + R 3
(14)
Definition 2: Equivalent resistance in a series circuit, RsRs

For nn resistors in series the equivalent resistance is:

R s = R 1 + R 2 + R 3 + + R n R s = R 1 + R 2 + R 3 + + R n
(15)

Let us apply this to the following circuit.

Figure 8
Figure 8 (PG11C9_008.png)

The resistors are in series, therefore:

R s = R 1 + R 2 + R 3 = 3 Ω + 10 Ω + 5 Ω = 18 Ω R s = R 1 + R 2 + R 3 = 3 Ω + 10 Ω + 5 Ω = 18 Ω
(16)
Exercise 2: Equivalent series resistance I

Two 10 kΩΩ resistors are connected in series. Calculate the equivalent resistance.

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

    Since the resistors are in series we can use:

    R s = R 1 + R 2 R s = R 1 + R 2
    (17)
  2. Step 2. Solve the problem :
    R s = R 1 + R 2 = 10 k Ω + 10 k Ω = 20 k Ω R s = R 1 + R 2 = 10 k Ω + 10 k Ω = 20 k Ω
    (18)
  3. Step 3. Write the final answer :

    The equivalent resistance of two 10 kΩΩ resistors connected in series is 20 kΩΩ.

Exercise 3: Equivalent series resistance II

Two resistors are connected in series. The equivalent resistance is 100 ΩΩ. If one resistor is 10 ΩΩ, calculate the value of the second resistor.

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

    Since the resistors are in series we can use:

    R s = R 1 + R 2 R s = R 1 + R 2
    (19)

    We are given the value of RsRs and R1R1.

  2. Step 2. Solve the problem :
    R s = R 1 + R 2 R 2 = R s - R 1 = 100 Ω - 10 Ω = 90 Ω R s = R 1 + R 2 R 2 = R s - R 1 = 100 Ω - 10 Ω = 90 Ω
    (20)
  3. Step 3. Write the final answer :

    The second resistor has a resistance of 90 ΩΩ.

Equivalent parallel resistance

Consider a circuit consisting of a single cell and three resistors that are connected in parallel.

Figure 9
Figure 9 (PG11C9_009.png)

The first principle to understand about parallel circuits is that the voltage is equal across all components in the circuit. This is because there are only two sets of electrically common points in a parallel circuit, and voltage measured between sets of common points must always be the same at any given time. So, for the circuit shown, the following is true:

V = V 1 = V 2 = V 3 V = V 1 = V 2 = V 3
(21)

The second principle for a parallel circuit is that all the currents through each resistor must add up to the total current in the circuit.

I = I 1 + I 2 + I 3 I = I 1 + I 2 + I 3
(22)

Also, from applying Ohm's Law to the entire circuit, we can write:

V = I R p V = I R p
(23)

where RpRp is the equivalent resistance in this parallel arrangement.

We are now ready to apply Ohm's Law to each resistor, to get:

V 1 = R 1 · I 1 V 2 = R 2 · I 2 V 3 = R 3 · I 3 V 1 = R 1 · I 1 V 2 = R 2 · I 2 V 3 = R 3 · I 3
(24)

This can be also written as:

I 1 = V 1 R 1 I 2 = V 2 R 2 I 3 = V 3 R 3 I 1 = V 1 R 1 I 2 = V 2 R 2 I 3 = V 3 R 3
(25)

Now we have:

I = I 1 + I 2 + I 3 V R p = V 1 R 1 + V 2 R 2 + V 3 R 3 = V R 1 + V R 2 + V R 3 because V = V 1 = V 2 = V 3 = V 1 R 1 + 1 R 2 + 1 R 3 1 R p = 1 R 1 + 1 R 2 + 1 R 3 I = I 1 + I 2 + I 3 V R p = V 1 R 1 + V 2 R 2 + V 3 R 3 = V R 1 + V R 2 + V R 3 because V = V 1 = V 2 = V 3 = V 1 R 1 + 1 R 2 + 1 R 3 1 R p = 1 R 1 + 1 R 2 + 1 R 3
(26)
Definition 3: Equivalent resistance in a parallel circuit, RpRp

For nn resistors in parallel, the equivalent resistance is:

1 R p = 1 R 1 + 1 R 2 + 1 R 3 + + 1 R n 1 R p = 1 R 1 + 1 R 2 + 1 R 3 + + 1 R n
(27)

Let us apply this formula to the following circuit.

Figure 10
Figure 10 (PG11C9_010.png)

What is the total resistance in the circuit?

1 R p = 1 R 1 + 1 R 2 + 1 R 3 = 1 10 Ω + 1 2 Ω + 1 1 Ω = 1 + 5 + 10 10 = 16 10 R p = 0 , 625 Ω 1 R p = 1 R 1 + 1 R 2 + 1 R 3 = 1 10 Ω + 1 2 Ω + 1 1 Ω = 1 + 5 + 10 10 = 16 10 R p = 0 , 625 Ω
(28)

Use of Ohm's Law in series and parallel Circuits

Exercise 4: Ohm's Law

Calculate the current (II) in this circuit if the resistors are both ohmic in nature.

Solution

Figure 11
Figure 11 (PG11C9_011.png)

  1. Step 1. Determine what is required :

    We are required to calculate the current flowing in the circuit.

  2. Step 2. Determine how to approach the problem :

    Since the resistors are Ohmic in nature, we can use Ohm's Law. There are however two resistors in the circuit and we need to find the total resistance.

  3. Step 3. Find total resistance in circuit :

    Since the resistors are connected in series, the total resistance RR is:

    R = R 1 + R 2 R = R 1 + R 2
    (29)

    Therefore,

    R = 2 + 4 = 6 Ω R = 2 + 4 = 6 Ω
    (30)
  4. Step 4. Apply Ohm's Law :
    V = R · I I = V R = 12 6 = 2 A V = R · I I = V R = 12 6 = 2 A
    (31)
  5. Step 5. Write the final answer :

    A 2 A current is flowing in the circuit.

Exercise 5: Ohm's Law I

Calculate the current (II) in this circuit if the resistors are both ohmic in nature.

Solution

Figure 12
Figure 12 (PG11C9_012.png)

  1. Step 1. Determine what is required :

    We are required to calculate the current flowing in the circuit.

  2. Step 2. Determine how to approach the problem :

    Since the resistors are Ohmic in nature, we can use Ohm's Law. There are however two resistors in the circuit and we need to find the total resistance.

  3. Step 3. Find total resistance in circuit :

    Since the resistors are connected in parallel, the total resistance RR is:

    1 R = 1 R 1 + 1 R 2 1 R = 1 R 1 + 1 R 2
    (32)

    Therefore,

    1 R = 1 R 1 + 1 R 2 = 1 2 + 1 4 = 2 + 1 4 = 3 4 T h e r e f o r e , R = 1 , 33 Ω 1 R = 1 R 1 + 1 R 2 = 1 2 + 1 4 = 2 + 1 4 = 3 4 T h e r e f o r e , R = 1 , 33 Ω
    (33)
  4. Step 4. Apply Ohm's Law :
    V = R · I I = V R = 12 4 3 = 9 A V = R · I I = V R = 12 4 3 = 9 A
    (34)
  5. Step 5. Write the final answer :

    A 9 A current is flowing in the circuit.

Exercise 6: Ohm's Law II

Two ohmic resistors (R1R1 and R2R2) are connected in series with a cell. Find the resistance of R2R2, given that the current flowing through R1R1 and R2R2 is 0,25 A and that the voltage across the cell is 1,5 V. R1R1=1 ΩΩ.

Solution
  1. Step 1. Draw the circuit and fill in all known values. :

    Figure 13
    Figure 13 (PG11C9_013.png)

  2. Step 2. Determine how to approach the problem. :

    We can use Ohm's Law to find the total resistance RR in the circuit, and then calculate the unknown resistance using:

    R = R 1 + R 2 R = R 1 + R 2
    (35)

    because it is in a series circuit.

  3. Step 3. Find the total resistance :
    V = R · I R = V I = 1 , 5 0 , 25 = 6 Ω V = R · I R = V I = 1 , 5 0 , 25 = 6 Ω
    (36)
  4. Step 4. Find the unknown resistance :

    We know that:

    R = 6 Ω R = 6 Ω
    (37)

    and that

    R 1 = 1 Ω R 1 = 1 Ω
    (38)

    Since

    R = R 1 + R 2 R = R 1 + R 2
    (39)
    R 2 = R - R 1 R 2 = R - R 1
    (40)

    Therefore,

    R 2 = 5 Ω R 2 = 5 Ω
    (41)

Batteries and internal resistance

Real batteries are made from materials which have resistance. This means that real batteries are not just sources of potential difference (voltage), but they also possess internal resistance. If the total voltage source is referred to as the emf, EE, then a real battery can be represented as an emf connected in series with a resistor rr. The internal resistance of the battery is represented by the symbol rr.

Figure 14
Figure 14 (PG11C9_014.png)

Definition 4: Load

The external resistance in the circuit is referred to as the load.

Suppose that the battery with emf EE and internal resistance rr supplies a current II through an external load resistor RR. Then the voltage drop across the load resistor is that supplied by the battery:

V = I · R V = I · R
(42)

Similarly, from Ohm's Law, the voltage drop across the internal resistance is:

V r = I · r V r = I · r
(43)

The voltage VV of the battery is related to its emf EE and internal resistance rr by:

E = V + I r ; o r V = E - I r E = V + I r ; o r V = E - I r
(44)

The emf of a battery is essentially constant because it only depends on the chemical reaction (that converts chemical energy into electrical energy) going on inside the battery. Therefore, we can see that the voltage across the terminals of the battery is dependent on the current drawn by the load. The higher the current, the lower the voltage across the terminals, because the emf is constant. By the same reasoning, the voltage only equals the emf when the current is very small.

The maximum current that can be drawn from a battery is limited by a critical value IcIc. At a current of IcIc, VV=0 V. Then, the equation becomes:

0 = E - I c r I c r = E I c = E r 0 = E - I c r I c r = E I c = E r
(45)

The maximum current that can be drawn from a battery is less than ErEr.

Exercise 7: Internal resistance

What is the internal resistance of a battery if its emf is 12 V and the voltage drop across its terminals is 10 V when a current of 4 A flows in the circuit when it is connected across a load?

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

    It is an internal resistance problem. So we use the equation:

    E = V + I r E = V + I r
    (46)
  2. Step 2. Solve the problem :
    E = V + I r 12 = 10 + 4 ( r ) = 0 . 5 E = V + I r 12 = 10 + 4 ( r ) = 0 . 5
    (47)
  3. Step 3. Write the final answer :

    The internal resistance of the resistor is 0.5 ΩΩ.

Resistance

  1. Calculate the equivalent resistance of:
    1. three 2 ΩΩ resistors in series;
    2. two 4 ΩΩ resistors in parallel;
    3. a 4 ΩΩ resistor in series with a 8 ΩΩ resistor;
    4. a 6 ΩΩ resistor in series with two resistors (4 ΩΩ and 2 ΩΩ ) in parallel.
  2. Calculate the total current in this circuit if both resistors are ohmic.
    Figure 15
    Figure 15 (PG11C9_015.png)
  3. Two ohmic resistors are connected in series. The resistance of the one resistor is 4 ΩΩ . What is the resistance of the other resistor if a current of 0,5 A flows through the resistors when they are connected to a voltage supply of 6 V.
  4. Describe what is meant by the internal resistance of a real battery.
  5. Explain why there is a difference between the emf and terminal voltage of a battery if the load (external resistance in the circuit) is comparable in size to the battery's internal resistance
  6. What is the internal resistance of a battery if its emf is 6 V and the voltage drop across its terminals is 5,8 V when a current of 0,5 A flows in the circuit when it is connected across a load?

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 16: An example of a series-parallel network. The dashed boxes indicate parallel sections of the circuit.
Figure 16 (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 16, 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 16 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 17
Figure 17 (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
(48)

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
(49)

Figure 18
Figure 18 (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
(50)

The equivalent resistance of the circuit in Figure 16 is 7,66ΩΩ.

Series and parallel networks

Determine the equivalent resistance of the following circuits:

  1. Figure 19
    Figure 19 (PG11C9_019.png)
  2. Figure 20
    Figure 20 (PG11C9_020.png)
  3. Figure 21
    Figure 21 (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 22
Figure 22 (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 8: Wheatstone bridge

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

Figure 23
Figure 23 (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
    (51)
  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
    (52)
  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 5: Electrical Power

Electrical power is calculated as:

P = I · V P = I · V
(53)

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
(54)
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
(55)

is identical to

P = I 2 R P = I 2 R
(56)

and

P = V 2 R P = V 2 R
(57)

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 24
    Figure 24 (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 25
    Figure 25 (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 3
    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 4
     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 26
    Figure 26 (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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