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Many modern technologies, such as computers and smartphones, are built on the principles of Physics Topics such as quantum mechanics and information theory.
What are the Resistance Rules in Parallel? What is Combination of Resistances?
Two or more resistors are said to be connected in parallel if one end of each is connected to a common point. and the other end to another point.
When this combination is joined in a circuit the main current distributed among the resistors but the potential difference across each is the same.
Calculation of equivalent resistance: Three resistances R1, R2, R3 are connected in parallel in between two common points A and B of an electrical circuit (Fig). Let VA and VB be the potentials at the points A and B respectively.
A current I is divided into three parts through these resistances.
Thus
I = I1 + I2 + I3 …… (1)
A and B are the common terminals of each of the three resistances. So according to Ohm’s law,
for the resistance R1, I1 = \(\frac{V_A-V_B}{R_1}\) ….. (2)
for the resistance R2, I2 = \(\frac{V_A-V_B}{R_2}\) …. (3)
for the resistance R3, I3 = \(\frac{V_A-V_B}{R_3}\) …. (4)
Adding (2), (3) and(4) weget,
I1 + I2 + I3 = (VA – VB)\(\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)\)
or, I = (VA – VB)\(\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)\) …. (5)
If R be the equivalent resistance of the combination and if it is connected between the points A and B, the main current flowing in the circuit will remain the same. So,
I = \(\frac{V_A-V_B}{R}\) …. (6)
From (5) and (6) we get, \(\frac{1}{R}\) = \(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
Similarly, if n number of resistances are connected in parallel instead of the three resistances, we have
\(\frac{1}{R}\) = \(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdots+\frac{1}{R_n}\) = \(\sum_{i=1}^n \frac{1}{R_i}\) …. (7)
So, reciprocal of the equivalent resistance of the parallel combination = sum of the reciprocals of the individual resistances.
Calculation of current in different resistances: If R he the equivalent resistance of a parallel combination connected between two points A and B of the circuit, the main current I is given by,
I = \(\frac{V_A-V_B}{R}\) or, VA – VB = IR
The potential difference of each resistance is (VA – VB).
So, current flowing through R1 is
I1 = \(\frac{V_A-V_B}{R_1}\) = \(\frac{I \cdot R}{R_1}\) = \(I \cdot \frac{R}{R_1}\)
This rule is applicable to each resistance of the parallel combination.
So, current flowing through any resistance
= main current \(\times \frac{\text { equivalent resistance }}{\text { corresponding resistance }}\)
A few characteristics of parallel combination of resistances:
i) The potential difference across each resistance is the same.
ii) \(\frac{1}{R}\) = \(\frac{1}{R_1}+\frac{1}{R_2}+\cdots\) ……. ; so \(\frac{1}{R}\) is greater than each of \(\frac{1}{R_1}\), \(\frac{1}{R_2}\),……etc. Therefore, the value of R is less than each of R1, R2, etc. i.e., the equivalent resistance is less than each of the resistances in the combination.
iii) Total current through the parallel combination is the sum of the individual currents through the resistances.
iv) Since, potential difference across each resistance is constant, individual currents are inversely proportional to the individual resistances.
In domestic electrical circuits, appliances such as bulb, electric fan, heater, etc. operate at the same potential difference (220 V). So these are connected in parallel.
Special case (parallel combination of two resistances): Suppose two resistances R1 and R2 are connected in parallel. Let their equivalent resistance be R. Then,
\(\frac{1}{R}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) or, R = \(\frac{R_1 R_2}{R_1+R_2}\)
Potential difference across the two ends of the combination,
V = IR = \(I \frac{R_1 R_2}{R_1+R_2}\)
Potential diffèrence of each resistance = V.
So, current flowing through R1, I1 = \(\frac{V}{R_1}\) = \(I \cdot \frac{R_2}{R_1+R_2}\)
Similarly, current flowing through R2, I2 = \(I \cdot \frac{R_1}{R_1+R_2}\)
So, current flowing through a resistance
= main current \(\times \frac{\text { other resistance }}{\text { sum of the two resistances }}\)
This formula will often be found handy in numerical calculations.
Numerical Examples
Example 1.
The equivalent resistance of two coils connected in series and in parallel are 12 Ω and \(\frac{5}{3}\)Ω respectively. Calculate the value of each resistance. [HS 03]
Solution:
If R1 and R2 be the two resistances, then equivalent resistance in series combination,
R1 + R2 = 12 ……. (1)
Equivalent resistance of parallel combination,
\(\frac{R_1 R_2}{R_1+R_2}\) = \(\frac{5}{3}\) or, \(\frac{R_1 R_2}{12}\) = \(\frac{5}{3}\) or, R1R2 = 20
Now, (R1 – R2)2 = (R1 + R2)2 – 4R1R2
= 122 – 4 × 20 = 144 – 80 = 64
or, R1 – R2 = 8 [Here we assume R1 > R2] ………. (2)
Adding equations (1) and (2) we have,
2R1 = 20 or, R1 = 10 Ω
From equation (1) we have,
R2 = 12 – R1 = 12 – 10 = 2Ω
Example 2.
A 5 ampere current Is distributed In three branches. The ratio of the lengths of the wires In the three branches is 1 : 2 : 3. DetermIne the magnitude of current in each branch. The material and the cross sectional area of each wire are the same.
Solution:
Resistance of the wire is proportional to its length as the material and the cross sectional area are the same. So we take the resistances as x, 2x and 3x and the equivalent resistance as R. We have,
\(\frac{1}{R}\) = \(\frac{1}{x}+\frac{1}{2 x}+\frac{1}{3 x}\) = \(\frac{6+3+2}{6 x}\) = \(\frac{11}{6 x}\) or R = \(\frac{6 x}{11}\)
So, potential difference between the two terminals,
V = IR = I ᐧ \(\frac{6 x}{11}\)
As the potential difference across each wire is the same, the current in the first branch
Example 3.
Current is allowed to pass in a circuit formed by two wires of the same material connected in a parallel combination. The ratio of the lengths and the radii of the two wires are 4 : 3 and 2: 3 respectively. Determine the ratio of the currents flowing through the two wires. [AIEEE ‘04]
Solution:
R = \(\rho \frac{l}{A}\) = \(\rho \frac{l}{\pi r^2}\)
So, \(\frac{R_1}{R_2}\) = \(\frac{l_1}{l_2} \cdot\left(\frac{r_2}{r_1}\right)^2\) = \(\frac{4}{3} \times\left(\frac{3}{2}\right)^2\) = 3
In parallel combination the current through a branch is inversely proportional to the resistance of the branch.
∴ \(\frac{I_1}{I_2}\) = \(\frac{R_2}{R_1}\) = \(\frac{1}{3}\)
∴ Ratio of currents flowing through the wires = I1 : I2 = 1 : 3.
Mixed Combination of Resistances
The complicated circuits of radio, television, etc. involve mixed combination of resistances.
Let us discuss the mixed combination with three resistances as a simple example.
Three resistances R1, R2, R3 are to be connected between two points A and B in a circuit in mixed combination. The resistance can be connected in the following ways.
i) The parallel combination of R1 and R2 is connected with
R3 in series (Fig. 1.11(a)j. The equivalent resistance of this combination,
R = \(\frac{R_1 R_2}{R_1+R_2}\) + R3
Changing the relative positions of R1 and R3 and again of R2 and R3, two more similar mixed combinations are obtained. Equivalent resistance changes accordingly.
ii) The series combination of R1 and R2 is connected with R3 in parallel (Fig.(b)). The equivalent resistance of the series combination of R1 and R2 is R’ = R1 + R2. Since R3 is connected in parallel with R’, the equivalent resistance of this mixed combination is
In this case also, by changing the relative positions of R1 and R3 and again of R2 and R3, two more similar mixed combinations are obtained. Equivalent resistance will change accordingly.
Numerical Examples
Example 1.
ABC is a triangle formed by wires. The resistances of the sides AB, BC and CA are respectively 40 Ω, 60 Ω and 100 Ω. What is the equivalent resistance between the points A and B?
Solution:
Along the path ACB the two resistances 1oo Ω and 60 Ω are connected in series [Fig.]. So the equivalent resistance of the path ACB = 100 + 60 = 160 Ω.
Again, the resistance 40 Ω of the side AB is connected in parallel with the path ACB. So the equivalent resistance between the two points A and B is
R = \(\frac{160 \times 40}{160+40}\) = \(\frac{160 \times 40}{200}\) = 32 Ω
Example 2.
Determine the equivalent resistance between the points A and B (Fig.).
Solution:
The equivalent resistance of the path ADC = 3 + 7 = 10 Ω. With the path ADC the 10 Ω resistance of the diagonal AC is connected in parallel.
So, equivalent resistance of AD, DC and AC
= \(\frac{10 \times 10}{10+10}\) = 5 Ω.
With this 5 Ω resistance, the 5 Ω resistance of the side BC is connected in series. So the equivalent resistance of 5 Ω and BC
= 5 + 5 = 10 Ω
Now, 1o Ω resistance of AB is also in parallel with the above equivalent resistance 10 Ω.
So, the equivalent resistance between A and B is
R = \(\frac{10 \times 10}{10+10}\) = 5 Ω
Example 3.
The two circuits In the Fig. draw equal currents from the battery. But the current through the resistance R In the second circuit is \(\frac{1}{10}\)th of that in the first circuit. Determine the values of R1 and R2.
Solution:
Current drawn from the battery in first case,
I1 = \(\frac{E}{R}\) …. (1)
Current drawn from the battery in the second case,
I2 \(=\frac{\text { total emf }}{\text { total resistance }}\) = \(\frac{E}{R_1+\frac{R R_2}{R+R_2}}\)
[∵ R1 is in series with the parallel combination of R and R2]
Required Resistance, R2 = 0.11R and R1 = 0.9R
Example 4.
You are given several identical resistances, each of value R = 10 Ω and each capable of carrying a maximum current of 1 A. It is required to make a suitable combination of these resistances to obtain a resistance of 5 Ω which can carry a current of 4 A. Find the minimum number of resistances of the type R that will be required.
Solution:
The resistances have to be connected in a series parallel combination. Since each resistance can carry current of 1 A, so to pass 4 A current, we need four paths in parallel.
Let r he the resistance of each path.
The equivalent resistance of 4 parallel paths will be \(\frac{r}{4}\)
According to the given problem,
\(\frac{r}{4}\) = 5
∴ r = 5 × 4 = 20 Ω
In order to have 20 Ω resistance in each path, two resistances
each of resistance 10 Ω have to be connected in series.
Since there are four paths, the total number of resistances required = 2 × 4 = 8.
Example 5.
A wire of uniform cross section and length l has a resistance of 16Ω. It is cut into four equal parts. Each part is stretched uniformly to length l and all the four stretched parts are connected In parallel. Calculate the total resistance of the combination so formed. Assume that stretching of wire does not cause any change in the density of its material.
Solution:
Let the cross section of the wire changes from A to A1 when it is cut into four parts.
As denstity remains constant, so volume also remains constant.
Volume of each part before stretching = volume of each part after stretching
or, \(\frac{A l}{4}\) = Al or, A1 = \(\frac{A}{4}\)
So resistance of each part, R1 = \(\frac{\rho l}{\frac{A}{4}}\) = \(\frac{4 \rho l}{A}\)
Therefore, equivalent resistance of the parallel combination,
Req = \(\frac{R_1}{4}\) = \(\frac{\rho l}{A}\) = 16Ω
[∵ initial resistance = \(\frac{\rho l}{A}\) = 16Ω]