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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 1
Figure 1 (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 2
Figure 2 (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
(1)

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
(2)

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
(3)

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
(4)

However, because

I = I 1 = I 2 = I 3 I = I 1 = I 2 = I 3
(5)

, 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 )
(6)

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

V = I · R V = I · R
(7)

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
(8)
Definition 1: 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
(9)

You can use the following simulation to test this result and all other results in this chapter.

Figure 3
Figure 3 (circuit-kit.png)
run demo

Let us apply this to the following circuit.

Figure 4
Figure 4 (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
(10)

Figure 5
Khan academy video on electric circuits - 1

Exercise 1: 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
    (11)
  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
    (12)
  3. Step 3. Write the final answer :

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

Exercise 2: 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
    (13)

    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
    (14)
  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 6
Figure 6 (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
(15)

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
(16)

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

V = I R p V = I R p
(17)

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
(18)

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
(19)

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
(20)
Definition 2: 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
(21)

Let us apply this formula to the following circuit.

Figure 7
Figure 7 (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
(22)

Figure 8
Khan academy video on electric circuits 2

Figure 9
Khan academy video on electric circuits 3

Use of Ohm's Law in series and parallel Circuits

Exercise 3: Ohm's Law

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

Solution

Figure 10
Figure 10 (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
    (23)

    Therefore,

    R = 2 + 4 = 6 Ω R = 2 + 4 = 6 Ω
    (24)
  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
    (25)
  5. Step 5. Write the final answer :

    A 2 A current is flowing in the circuit.

Exercise 4: Ohm's Law I

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

Solution

Figure 11
Figure 11 (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
    (26)

    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
    (27)
  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
    (28)
  5. Step 5. Write the final answer :

    A 9 A current is flowing in the circuit.

Exercise 5: 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 12
    Figure 12 (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
    (29)

    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
    (30)
  4. Step 4. Find the unknown resistance :

    We know that:

    R = 6 Ω R = 6 Ω
    (31)

    and that

    R 1 = 1 Ω R 1 = 1 Ω
    (32)

    Since

    R = R 1 + R 2 R = R 1 + R 2
    (33)
    R 2 = R - R 1 R 2 = R - R 1
    (34)

    Therefore,

    R 2 = 5 Ω R 2 = 5 Ω
    (35)

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 13
Figure 13 (PG11C9_014.png)

Definition 3: 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
(36)

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

V r = I · r V r = I · r
(37)

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
(38)

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
(39)

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

Exercise 6: 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
    (40)
  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
    (41)
  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 14
    Figure 14 (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?

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