Devices in Series and Parallel

Concept Explanation

Devices in Series and Parallel

Sometimes we come across a circuit which consists of more than one type of combination of resistance. That means some resistance may be parallel to each other and this combination can be in series with another resistance. Or we have different parallel combinations of resistance which are in series to each other. A combination circuit has some resistances connected in series combination and some parallel combination.

This type of combination is also called complex circuit. While solving problems of mixed combination of resistances, there are many points to be considered as given below:

The first step is to reduce the mixed combination of resistance into a simple circuit containing only one resistor. For this examine, the given circuit and replace the resistors that are connected in parallel or in series with their equivalent resistances.

Draw the new circuit after making the changes and repeat the same procedure again and again till a simple circuit is obtained.

If the current through or potential difference across a resistor in the complex circuit is to be found, then start with the simple circuit reduced from the complex circuit and gradually work your way back through the circuits, using V= IR.

While calculating the equivalent resistance, do not consider the battery as we consider that the battery has negligible resistance. But, if its resistance is given, then consider it as an individual resistor.

Illustration: Consider the circuit diagram as shown in the figure

$large If ;R_{1}=R_{2}=R_{3}=R_{4}=R_{5}=3Omega$

then find the equivalent resistance of the circuit.

Solution: From the combination, it can be observed that $dpi{100} large R_{2};and;R_{3}$ are connected in series order. As current through $large R_{2};and;R_{3}$ is same. So, their equivalent resistance is

$R^i=R_{2}+R_{3}=3Omega +3Omega =6Omega$

Now, the given circuit can be redrawn as shown in the figure

Now, it can be seen that $large R_{4};and;R^i$ are in parallel combination. As, currents through $large R_{4};and;R^i$ are different. So, their equivalent resistance can be calculated as below

$frac{1}{R^{ii}}= frac{1}{R_4}+ frac{1}{R^{i}}=frac{1}{3}+frac{1}{6}=frac{2+1}{6}=frac{3}{6}=frac{1}{2}$

$therefore ;R^{ii}=2;Omega$

Now, the given circuit can be redrawn as shown in the figure

Now, it is clear from the diagram that all the resistances $large R_{5},R^{ii};and;R_{1}$ are in series combination.

As, current through $large R_{1},R^{ii};and;R_{5}$ is same.

$large therefore$ Equivalent resistance of the circuit is

$large R=R_{5}+R^{ii}+R_{1}=3Omega +2Omega +3Omega =8Omega$

The battery has potential as = 16 V

By using the formula V = IR

$I=frac{V}{R}= frac{16}{8}= 2 A$

As current flowing through $large R_{1},R^{ii};and;R_{5}$ is same because they are in series. So

$I_1= 2 A;,;I^{ii}= 2 A; and; I_5= 2 A$ and potential drop across them is

$V_1= 2; X ;3= 6;V ;,;V^{ii}= 2; X; 2= 4;V; and; V_5= 2; X;3= 6; V$

As $R^{ii}$ is an equivalent resistance for a parallel combination of $large R_{4};and;R^i$ , So potential drop across them is the same

$I_4= frac{V_4}{R_4}=frac{4}{3}=1.33;A; and; I^i= frac{V^i}{R^i}=frac{4}{6}=0.67;A$

As $R^i$ is an equivalent resistance for a series combination of $large R_{2};and;R_3$ , So current flowing through is the same i.e 0.67 A

$V_2= frac{2}{3}; X ;3= 2;V; and; V_3= frac{2}{3}; X;3= 2; V$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

Consider the following statement: (A) In series connection, the same current flows through each element. (B) In parallel connection, the same potential difference gets applied across each element.
A. Both A and B are correct	B. A is correct but B is wrong	C. A is wrong but B is correct	D. Both A and B are wrong
Right Option : A
View Explanation

Question : 2

Marie conducted an experiment to find out if the number of bulbs arranged in series would affect the brightness of the bulbs. which of the following variables should be kept the same? A. Type of bulbs B. number of wires C. Types of batteries D. number of bulbs
A. D only	B. B and D only	C. A,B and C only	D. A,B,C and D
Right Option : C
View Explanation

Question : 3

Calculate the equivalent resistance between A and B
A. $6.67;Omega$	B. $46.76 ;Omega$	C. $26.67;Omega$	D. $10.67:Omega$
Right Option : C
View Explanation

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Devices in Series and Parallel

Devices in Series and Parallel

Students / Parents Reviews [10]

Prisha Gupta

Yatharthi Sharma

Cheshta

Barkha Arora

Eshan Arora

Shreya Shrivastava

Archit Segal

Shivam Rana

Manish Kumar

Aman Kumar Shrivastava