Chapter 8.6 - Comparison between Series and parallel Resistors

In the section 8.4, we saw series connection of resistors. In the previous section 8.5 we saw parallel connection of resistors. In this section, we will see a comparison between the two. We will write the comparison in steps:

1. Consider fig.8.25(a) below:

Fig.8.25

• People are crossing 3 canals. For that purpose, 3 bridges are provided. 
2. Because of the necessity to use the bridges, the movement of the people is somewhat constrained.
• We can compare this arrangement to resistors in series. When resistors are provided, free flow of current is constrained. 
3. If any one of the bridges in series is broken down, the movement of people will no longer be possible. This is shown in fig.b. 
• If in our home wiring, a TV, refrigerator and electric fan are connected in series, we will have to turn all the three of them on. 
• If any one is turned off, the circuit will not complete and so the other devices will also be turned off  
4. Now consider fig.8.26(a) below:

Fig.8.26

• People are crossing a canal. 3 bridges are provided for the purpose. 
• People can use any one of the three bridges. 
• So there is greater freedom of movement. 
5. This arrangement can be compared to resistors in parallel. 
• Even if one of the bridges is broken down, people can take the other bridges. 
• The devices in our home are connected in parallel circuits. 
• So even if one or two devices are turned off, the other devices will work.

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Let us do an activity to compare series and parallel resistors:
• In this activity, we have to make two circuits. They are shown in figs.8.27 (a) and (b). 
• We can note the peculiarities of each:
Circuit 1: There are two 6 W, 6 V bulbs, a 6 V battery and a switch
    ♦ The two bulbs are connected in parallel
Circuit 2: Components are the same: Two 6 W, 6 V bulbs, a 6 V battery and a switch
    ♦ The two bulbs are connected in series
• Now we can begin the trials:
Trial 1: Turn both switches on. This is shown in figs.8.27(a) and (b)

Fig.8.27

Observations: 
(i) The two bulbs in circuit 1 glow with equal intensity. Let this intensity be C₁T₁ 
['C₁' for circuit 1 and 'T₁' for trial 1]
(ii) The two bulbs in circuit 2 glow with equal intensity. Let this intensity be C₂T₁
(iii) The intensity in circuit 1 is more. That is., C₁T₁ > C₂T₁
• Before proceeding to trial 2, we will analyse the above observations. We will write the analysis in steps:
1. We see that, bulbs connected in series are dim. 
• We want to know the reason.
2. All bulbs are of the same type. So they have the same resistance. Let this be 'r'
• Let us calculate the effective resistance 'R' in circuit 1:
• The resistors are connected in parallel.
• We have: ¹⁄_R = (¹⁄_r + ¹⁄_r) = ²⁄_r  ⟹ R = ^r⁄₂
• So total current I = ^V⁄_R = ^2V⁄_r
3. At the junction, this current will branch into two: I₁ and I₂
We have:
• I₁ = ^V1⁄_r
• I₂ = ^V2⁄_r
4. But in parallel connection, V₁ = V₂ = V
• So we get: I₁ = I₂ = ^V⁄_r = ⁶⁄_r
• Both the currents are the same. So the two bulbs glow with equal intensity C₁T₁
5. Let us calculate the effective resistance in circuit 2:
• The resistors are connected in series
• We have: R = (r+r) = 2r
6. Total current in the circuit = ^V⁄_2r
• This current will flow through the two bulbs. 
• So current in each of the two bulbs in series = ^V⁄_2r = ⁶⁄_2r = ³⁄_r
• Since both bulbs have the same current, they both glow with the same intensity C₂T₁
7. Now let us compare the two systems:
• Current for each bulb in the parallel connection = ⁶⁄_r
• Current for each bulb in the series connection = ³⁄_r
• Obviously, more current is passing through the bulbs in parallel connection. 
■ So they glow with greater intensity. That is: C₁T₁ > C₂T₁.

Trial 2: Remove one bulb from each circuit. This is shown in figs.8.28 (a) and (b) below:

Fig.8.28

• Note that, when a bulb is removed from circuit 2, alternate path has to be given. Other wise the circuit will not become complete
• But no such alternate path is required for circuit 1
• Turn both the switches on
Observations:
(i) Let the intensity of light in circuit 1 be C₁T₂. 
• We see that this intensity is same as what was seen in trial 1. That is., C₁T₂ = C₁T₁
(ii) Let the intensity of light in circuit 2 be C₂T₂. 
• We see that this intensity is more than what was seen in trial 1. That is., C₂T₂ > C₂T₁
• Also we see that this intensity in circuit 2 is same as that in circuit 1 in trial 1
That is., C₂T₂ = C₁T₂ = C₁T₁ 
1. Let us calculate the effective resistance in circuit 1:
• There is only one bulb. So the effective resistance is r
• Current I = ^V⁄_r = ⁶⁄_r
• Note that, this is the same current that each of the bulbs in circuit 1 had, in trial 1. 
• So the intensity is same.
2. Let us calculate the effective resistance in circuit 2:
• There is only one bulb. So the effective resistance is r
• Current I = ^V⁄_r = ⁶⁄_r
• Note that, this is the same current that each of the bulbs in circuit 1 had, in trial 1. 
• So the intensity is same.

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A virtual lab to make different circuits can be seen here.

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Now we will see some solved examples based on what we have learned so far in this chapter:
Solved example 8.1
A resistor has a length of 1m. It's area of cross section is 1 m². It's resistance value is R Ω.
Then answer the following questions:
(i) If the area is kept at 1 m² and the length is doubled, what will be the resistance?
(ii) If the length is kept at 1 m and area is doubled, what will be the resistance?
(iii) If both length and area are doubled, what will be the resistance?
(iv) If the length is kept at 1 m and area is halved, what will be the resistance?
Solution:
• We have: R = ρ × ^l ⁄_A
• Given l = 1 m, A = 1 m². 
• Substituting in the formula for R, we get: R = ρ × ¹ ⁄₁ = ρ Ω.
■ So the resistance value 'R' of the original resistor is equal to ρ. That is., R = ρ.

Part (i):
• Given l = 2 m, A = 1 m². 
• Let the new resistance value be R₁ 
• We get: R₁ = ρ × ² ⁄₁ = (2ρ) Ω = (2R) Ω. [∵ ρ = R] 
• So we can write: The new resistance is twice the original value
Part (ii):
• Given l = 1 m, A = 2 m². 
• Let the new resistance value be R₂ 
• We get: R₂ = ρ × ¹ ⁄₂ = (0.5ρ) Ω = (0.5R) Ω. [∵ ρ = R] 
• So we can write: The new resistance is half the original value
Part (iii):
• Given l = 2 m, A = 1 m². 
• Let the new resistance value be R₃ 
• We get: R₃ = ρ × ² ⁄₂ = (ρ) Ω = (R) Ω. [∵ ρ = R] 
• So we can write: The new resistance same as the original value
Part (iv):
• Given l = 1 m, A = 0.5 m². 
• Let the new resistance value be R₄ 
• We get: R₄ = ρ × ¹ ⁄_0.5 = (2ρ) Ω = (2R) Ω. [∵ ρ = R] 
• So we can write: The new resistance is twice the original value

Solved example 8.2
What is the current if 4 Ω and 2 Ω resistors are connected in series and 6 V potential difference is applied?
Solution:
1. When resistors are connected in series, the effective resistance R is given by:
R = (R₁+ R₂ + . . . + R_n)
• So in our present case, R = (4+2) = 6 Ω.
2. We have: I = ^V⁄_R
• Substituting the values, we get: I = ⁶⁄₆ = 1 A

Solved example 8.3
What is the current if 12 Ω and 4 Ω resistors are connected in parallel and 12 V potential difference is applied?
Solution:
1. When resistors are connected in parallel, the effective resistance R is given by:
R = (¹⁄_R1 + ¹⁄_R2+ ¹⁄_R3 + . . . + ¹⁄_Rn)
• So in our present case, ¹⁄_R = (¹⁄₁₂ + ¹⁄₄) = ⁴⁄₁₂ = ¹⁄₃
⟹ R = 3 Ω
2. We have: I = ^V⁄_R
• Substituting the values, we get: I = ¹²⁄₃ = 4 A

Solved example 8.4
(a) 10 resistors of 2 Ω each are connected in parallel. Calculate the effective resistance.
(b) 10 resistors of 2 Ω each are connected in series. Calculate the effective resistance.
Solution:
Part (a):
• When resistors having the same resistance value are connected in parallel, the effective resistance is given by: R = ^r⁄_n
• Where R is the effective resistance and r is the resistance of individual resistors 
• So we get: R = ²⁄₁₀ = ¹⁄₅ Ω.
Part (b):
• When resistors having the same resistance value are connected in series, the effective resistance is given by: R = nr
• Where R is the effective resistance and r is the resistance of individual resistors 
• So we get: R = 20 × 2 = 20 Ω.

Solved example 8.5
The resistance of a 10 cm long wire is 12 Ω. If this is folded into two parts of equal length and included in a circuit, how much will be the resistance produced?
Solution:
1. We have: R = ρ × ^l ⁄_A
• Given R = 12 Ω, l = 10 cm = 0.1 m 
2. Substituting in the formula for R, we get: 12 = ρ × ^0.1 ⁄_A 
⟹ ¹² ⁄_0.1 = ρ × ¹ ⁄_A 
⟹ ρ × ¹ ⁄_A = 120
3. Now the wire is folded into two equal parts. So we get:
• New length l1 = 5 cm = 0.05 m
• New area A1 = 2A
• Let the new resistance be R1
4. Substituting in the formula for R, we get: R1 = ρ × ^0.05 ⁄_2A .
⟹ ^R1 ⁄_0.05 = ρ × ¹ ⁄_2A  ⟹ ^2R1 ⁄_0.05 = ρ × ¹ ⁄_A 
⟹ 40R1 = ρ × ¹ ⁄_A.
5. But from (2), we have: ρ × ¹ ⁄_A = 120
So (4) becomes:
40R1 = 120 ⟹ R1 = 3 Ω.

Solved example 8.6
(a) A potential difference of 6 V is applied across a conductor having 12 Ω resistance. How much current will pass through it?
(b) How many times will the current increase if the length of the resistor is halved and potential difference is doubled?
Solution:
Part (a):
1. The given values are: V = 6 V, R = 12 Ω
We have to calculate I
2. We have: I = ^V⁄_R 
Substituting the known values, we get: I = ⁶⁄₁₂ = 0.5 A

Part (b):
1. The length of the resistor is halved. We have to find the new resistance.
• Let l be the original length and A the original area of cross section
• Then original resistance R = 12 = ρ × ^l ⁄_A ⟹ ρ = 12 × (^A ⁄_l)
2. Let l1 be the new length. Given that l1 = 0.5l
• Area of cross section does not change. So it is A
• So new resistance R1 = ρ × ^l1 ⁄_A = ρ × ^0.5l ⁄_A 
⟹ R1 = 0.5 × ρ × (^l ⁄_A)
3. From (1), we have: ρ = 12 × (^A ⁄_l)
• Substituting this in 2, we get:
R1 = 0.5 × [12 × (^A ⁄_l)] × (^l ⁄_A) = 0.5 × 12 = 6 Ω.
4. The voltage is doubled. So the new voltage is 12 V
So new current = I1 = ^V⁄_R= ¹²⁄₆ = 2 A
5. We have: ^I1⁄_I = ²⁄_0.5 = 4
• So the current increases 4 times.

Solved example 8.7
What mode of connection of three resistors of 3 Ω each will produce minimum resistance? What will produce maximum resistance? Draw the diagram and calculate the effective resistance
Solution:
• The two modes possible are: Series and parallel. They are shown in figs.8.29 (a) and (b) respectively

Fig.8.29

• The effective resistance in series connection is given by R = nr (Details here)
    ♦ So for our present problem, R = 3 × 3 = 9 Ω
• The effective resistance in series connection is given by R = ^r⁄_n (Details here)
    ♦ So for our present problem, R = ³⁄₃ = 1 Ω.
• So we can write:
    ♦ Series mode creates maximum resistance
    ♦ Parallel mode creates minimum resistance

Solved example 8.8
Draw a circuit having a battery, 3 bulbs, a switch and rheostat. In how many ways can the bulbs be connected? When is more light obtained?
Solution:
• The four different possibilities are shown in fig.8.30 below:
    ♦ In all the cases, the voltage available from the battery is the same. Let it be V
    ♦ In all the cases, the resistance of each bulb is the same. Let it be r

Fig.8.30

Fig.8.30(a):
1. Effective resistance = nr = 3r
2. So current released from the battery = I = ^V⁄_R = ^V⁄_3r
• Each of the 3 bulbs gets this current 
Fig.8.30(b):
1. Effective resistance = ^r⁄_n = ^r⁄₃
2. So current released from the battery = I = ^V⁄_R = ^3V⁄_r
3. At the junction, this current will split into 3 equal parts because, resistance of all bulbs are the same.
4. So each of the 3 bulbs gets ^V⁄_r
Fig.8.30(c):
1. Effective resistance = (nr + ^r⁄_n) = (r + ^r⁄₂ ) = ^3r⁄₂ 
2. So current released from the battery = I = ^V⁄_R = ^2V⁄_3r
• The first bulb gets this current. 
3. The two bulbs in parallel will get half of this current, which is equal to ^V⁄_3r.
Fig.8.30(d):
1. Inside the parallel mode, two bulbs are in series. Their effective resistance = 2r
2. So the effective resistance is given by: ¹⁄_R = (¹⁄_R1 + ¹⁄_R2)
⟹ ¹⁄_R = (¹⁄_2r + ¹⁄_r) = ³⁄_2r. 
• Effective resistance = R = ^2r⁄₃
3. So current released from the battery = I = ^V⁄_R = ^3V⁄_2r
4. Current in the upper branch = ^V⁄_{resistance of upper branch} = ^V⁄_2r
5. Current in the lower branch = ^V⁄_{resistance of lower branch} = ^V⁄_r
■ Conclusion:
• The currents in different bulbs are:
^V⁄_3r , ^V⁄_r , ^2V⁄_3r , ^V⁄_3r , ^V⁄_2r ,^V⁄_r
• The highest value is ^V⁄_r . The two cases below has this current. They glow with maximum light
    ♦ All bulbs in fig.(b)
    ♦ The bottom bulb in fig.(d)
• The lowest value is ^V⁄_3r . The two cases below has this current. They glow with least light
    ♦ All bulbs in fig.(a)
    ♦ The two parallel bulbs in fig.(d)

Solved example 8.9
In a circuit, 3 cells are connected as shown in fig.8.31 below. What is the resultant voltage?

Fig.8.31

Solution:
• The two cells in parallel will give an effective voltage of 1.5 V
• So the net effective voltage is (1.5 + 1.5) = 3 V

Solved example 8.10
Find the effective resistance in the three circuits shown in fig.8.32 below:

Fig.8.32

Solution:
Fig.8.32(a):
• Effective resistance = nr = 4 × 2 = 8 Ω.
Fig.8.32(b):
• Effective resistance in upper branch = nr = 2 × 2 = 4 Ω
• Effective resistance in lower branch = nr = 2 × 2 = 4 Ω
• Net effect = ^r⁄_n = ⁴⁄₂ = 2 Ω
Fig.8.32(c):
• Net effect of the parallel resistors = ^r⁄_n = ²⁄₂ = 1 Ω.
• Net effect of the whole circuit = (2+2+1) = 5 Ω

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In the next chapter, we will see electricity and magnetism.

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Chapter 5.1 - Wave length, Frequency and Time period

In the previous section we saw the basic details about Longitudinal waves. In this section we will see some mathematical calculations involved with it.

Consider the compressions and rarefactions produced by a sound source shown in the fig.5.3(a) below:

Fig.5.3

It is clear that the molecules between A and M are under various levels of pressure.  For example:
• Molecules that fall along the magenta coloured lines will be experiencing maximum compression. That is., maximum pressure
• Molecules that fall along the yellow coloured lines will be experiencing the least compression. That is., least pressure
• Other molecules which fall along neither of these lines will be experiencing intermediate pressures. 
• Those which are closer to the magenta lines will be experiencing more pressure
• Those which are closer to the yellow lines will be experiencing less pressure
• So we find that, there is a variation of pressure along the path from the sound source to the ear
■ We can plot a graph with distance on the x-axis and pressure on the y-axis. Let us see how the plotting is done:
1. Consider the left most magenta line. Get the pressure experienced by the molecules along that line. Mark this pressure on the Y axis. This is point A.  
2. Consider the left most yellow line. Get the pressure experienced by the molecules along that line.
    ♦ Mark this pressure on the Y axis. 
    ♦ Draw a horizontal yellow dotted line through this mark
    ♦ Draw a vertical yellow dotted line from the left most yellow line
    ♦ The horizontal and vertical yellow dashed lines will intersect at a point C
    ♦ The x coordinate of C will give the distance from the sound source and the y coordinate will give the pressure.
3. So we successfully marked two points A and C 
• y coordinate of A will give the pressure experienced by all molecules which fall along the left most magenta line
    ♦ x coordinate of A will be zero because, it's distance from the sound source is zero
• y coordinate of C will give the pressure experienced by all molecules which fall along the left most yellow line
    ♦ x coordinate of C will be the distance from the sound source
4. The intermediate points between A and C can be marked in the same way. 
    ♦ y coordinate of those points will be the pressure, 
    ♦ x coordinates will be the distance from the sound source. 
5. Once we mark those points, we can join them. We will get a smooth curve between A and C 
6. In this way, we can mark several points between C and M. 
    ♦ A sample point at E is shown in the fig.5.3(a)
7. When we join all such points, we will get a smooth curve between A and M. 
■ This curve is the graphical representation of the sound produced by the tuning fork. 
8. We have to make a modification to this graph:
• Consider the pressure value at A. It is the maximum compression value
• Consider the pressure value at C. It is the least compression value
• Take the average of the two. From maths classes we know that, average of the two will be the exact midpoint between the following two points:
    ♦ The point A
    ♦ Projection of C on the y-axis. That is., the point of intersection of the horizontal yellow dotted line and the y axis
9. Mark this average point on the y axis.
10. Draw the x-axis through this average point. This is shown in fig.5.3(b) 
    ♦ All points on the graph above the new x-axis represents a compression 
    ♦ All points below the new x-axis represents a rarefaction.
■ This is the final graphical representation.
• Points B, D, F, H, J and L are the intersection points between the graph and the new x-axis.
• If we draw vertical lines through these points, the molecules which fall along those points will be experiencing the average pressure  

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Let us learn the salient features of this graph. The graph in fig.5.3(b) is shown again in fig.5.4 below:

Fig.5.4

• We can see that it contains some peak points and some valley points. 

• The peak points are called crests of the wave. A, E, I and M are crests. 

• Valley points are called troughs of the wave. C, G and K are troughs.

• The distance between two consecutive crests is called the wavelength. It is represented by λ (Greek letter lambda). The unit of wavelength is m. 

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Next we have to find the frequency of the wave. Let us see how it is calculated:

1. We have seen that the sound is propagated as compressions (represented by crests in the graph) and rarefactions (represented by troughs in the graph).
2. Suppose that we are standing near the path of the wave, ready with a stop watch. Let the point where we are standing be 'P'
3. At the instant when a compression pass the point P, start the stop watch
4. Count the 'number of compressions' n, that pass the point P for any convenient time interval t, say 10 or 15 seconds. 
5. Then the ratio ⁿ⁄_t will give the 'number of compressions' that pass the point p in one second.
6. We know that, frequency is the 'number of occurrences of an event in unit time'
So ⁿ⁄_t is the frequency of the wave. It is represented by ν (Greek letter nu). It's unit is hertz (symbol Hz).
• This name is given in honour of the German scientist Heinrich Rudolph Hertz, whose works laid the foundation of the future developments of radio, telegraph, telephone etc.,    

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Let us repeat the experiment. This time, to determine another quantity.

1. Suppose that we are standing near the path of the wave, ready with a stop watch. Let the point where we are standing be 'P'
2. At the instant when a compression pass the point P, start the stop watch
3. At the instant when the next compression pass the point P, stop the stop watch. Note down the time 'T'
4. This is the time taken between two consecutive compressions. It is called the 'time period' of the wave. It is represented by T. It's unit is seconds.

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In the first experiment also, we did a similar procedure. Is there any inter connection?

Let us see:
1. In the second experiment, we cannot take any convenient time
• We must stop the stop watch exactly at the instant at which the next compression pass the point P
2. So applying the result from the first experiment into the second, we have t = T
• Also, only one compression passes in a time of 'T'. So 'n' = 1
3. Thus we get: ν = ⁿ⁄_t = ¹⁄_T
• We can write:
Eq.5.1: ν = ¹⁄_T
4. That means, frequency of a wave is the reciprocal of it's time period
5. ¹⁄_T has the unit s^-1. So ν also has the unit s^-1
6. That means, we can use either of the two units s^-1 or Hz for the frequency ν 

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Speed of a wave

We know how to find the speed of a car. Let us recall:
1. Mark two points at a convenient distance s apart. Say 150 or 250 m.
2. Note the time t₁ when the car passes the first point 
3. Note the time t₂ when the car passes the second point
4. The difference (t₂ – t₁) will give the time t required by the car to travel s
5. So speed of the car = ^distance⁄_time = ^s⁄_t
■ The same method can be used to determine the speed of a train. 
A train is a very long object. But we can take the times when the 'train's front end' passes the two points. The calculations are same as that of the car.

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■ For the train, another method can also be used:

1. Consider fig.5.5 below. It shows a train moving from left to right. 

Fig.5.5

2. Two red arrows A and B are marked at the midpoint of two adjacent compartments. Let the distance between those two arrows be s. A green arrow C is marked on the platform.
3. Note the time t₁ when the arrow B passes the arrow C. Note the time t₂ when the arrow A passes the arrow C
4. The difference (t₂ – t₁) will give the time t required by the train to travel s
5. So speed of the train =  ^distance⁄_time = ^s⁄_t

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The same method can be used to find the speed of a wave. 

1. The two arrows A and B correspond to two consecutive crests of a wave
2. Then the distance s will correspond to λ
3. We know that, the time required for two consecutive crests to pass a fixed point such as the green arrow C is the time period T
4. So speed v of the wave can be obtained as: v =  ^distance⁄_time = ^λ⁄_T
5. This can be written in another form also:
v = ^λ⁄_T ⇒ v = λ × ¹⁄_T ⇒ v = λ × ν (∵ ¹⁄_T = ν)
We can write it in the form of an equation:
Eq.5.2:
v = λ × ν
That is., Speed of sound = wavelength × frequency

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Now we will see a solved example

Solved example 5.1
A sound wave has a frequency of 2 kHz and a wavelength of 35 cm. How long will it take to travel 1.4 km?
Solution:
■ We have, speed = wavelength × frequency 
Given:
• Wave length, λ = 35 cm = 0.35 m
• Frequency, ν = 2 kHz = 2000 Hz = 2000 s^-1  
• So we get: v = 0.35 (m) × 2000 (s^-1) = 700 m s^-1  
■ We have, speed, v = ^distance⁄_time ⇒ time = ^distance⁄_speed
Given:
• Distance = 1.4 km = 1400 m
• So time required to travel 1.4 km = ¹⁴⁰⁰⁄₇₀₀ = 2 s 

Solved example 5.2
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m s^-1 in a given medium
Solution:
■ We have, speed, v = λ × ν
So we can write: 440 = λ × 220 ⇒ λ = ⁴⁴⁰⁄₂₂₀ = 2 m

Solved example 5.3
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?  
Solution:
■ We have seen that the time interval between two consecutive compressions or rarefactions is the time period T of the wave. So in this problem we have to calculate T
■ We have also seen that frequency is the reciprocal of T. That is., ν = ¹⁄_T (Eq.5.1)
⇒ T = ¹⁄ν 
• So we get T = ¹⁄₅₀₀ = 0.002 seconds

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In the next section, we will see some more properties of sound waves. 

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