Cells in Series and Parallel Important Questions

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Cells in series and parallel are the two ways of connection in a circuit. This topic has more weightage in Class 12 Chapter 3 Current Electricity. In a series connection, the cells are connected end to end whereas in a parallel connection, the cells are connected in parallel to one another. 

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Series and Parallel Circuit

Series and Parallel Circuit


Advantage of Parallel connection over Series connection

  • If any of the electrical components in a series connection gets disconnected, the entire circuit fails.
  • But this is overcome in a parallel circuit, where even if any of the components is broken, the rest of the circuit continues working.
  • This is why parallel circuits are used for domestic purposes so that even if any electrical device is disconnected, others continue to work.
  • In a parallel circuit, the internal resistance of the circuit is very minimal compared to the series combination.

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Important Questions on Combinations of cells in Series and Parallel

Ques. Batteries of 10V and 5 V are connected in series such that their emfs point in the same direction. Find the equivalent resistance for the system. (2 Marks)

Ans. The formula for equivalent series emf is given by,

Eeq = E1 + E2 + … 

Given

  • E1 = 10 V 
  • E2 = 5 V

Substituting these values in the equation, 

E = E1 + E2 

E = 10 + 5

E = 15 V

Ques. Batteries of 3, 5, and 10 ohms are connected in series such that their emfs point in the same direction. Find the equivalent resistance for the system. (2 Marks)

Ans. The formula for equivalent series emf is given by,

Eeq = E1 + E2 + E3 … 

Given

  • E1 = 3 V
  • E2 = 5 V
  • E3 = 10 V

Substituting these values in the equation, 

E = E1 + E2 + E3

E = 3 + 5 + 10 

E = 18 V

Ques. Batteries of 10V and 5 V are connected in series such that their emfs point in the same direction. The internal resistances of the batteries are 2 and 10 ohms respectively. Find the equivalent resistance for the system. (3 Marks)

Ans. The formula for equivalent series emf is given by,

Eeq = E1 + E2 + E3 … 

Given

  • E1 = 10 V
  • E2 = 5 V

Substituting these values in the equation, 

E = E1 + E2 

E = 10 + 5

E = 15 V

Equivalent resistance is also given by a similar equation, 

req = r1 + r2 + ….

  • r1 = 2 ohm
  • r2 = 10 ohm

substituting these values in the equation, 

r = r1 + r2 

⇒ r = 2 + 10

⇒ r = 12 ohms

Ques. Three batteries of internal resistances 2, 2, and 4 ohms are connected in parallel. Find the equivalent resistance for the system. (3 Marks)

Ans. Given

  • R1 = 2 ohm
  • R2 = 2 ohm
  • R3 = 4 ohm

The formula for equivalent resistance is given by,

\(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ......\)

Substituting these values in the equation, 

1/R = 1/2 + 1/2 + 1/4

1/R = (2+2+1)/4

1/R = 5/4

R = 4/5 ohm

Ques. Three batteries of internal resistances 5 ohm, 5 ohm, and 10 ohm, each of 10 V are connected in parallel. Find the equivalent resistance and emf for the system. (3 Marks)

Ans. 

Given

  • R1 = 5 ohm
  • R2 = 5 ohm
  • R3 = 10 ohm

The formula for equivalent resistance is given by,

\(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ......\)

Substituting these values in the equation, 

1/R = 1/5 + 1/5 + 1/10

1/R = (2+2+1)/10

1/R = 5/10 = 1/2

R = 2 ohm

Now when n number of identical batteries are connected in parallel then equivalent emf is equal to the emf due to a single cell. Therefore

Eequivalent = 10 V

Ques. Why is it preferable to connect bulbs in parallel than in series? (2 Marks)

Ans. Bulbs are connected in parallel so that even if one of the bulbs is disconnected, the others continue to get a current supply.

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Ques. Calculate the total resistance between points A and B. (3 Marks)

Calculate the total resistance between points A and B

Ans. From figure, the equivalent resistance of 1 ohm and 2 ohm resistance is given vy

Equivalent resistance, 1/Req = 1/R1 + 1/R2 +....

1/Req = 1/1 + 1/2 

1/Req = (2+1)/2

1/Req = 3/2

Req = 2/3 

Total Resistance of the circuit = 4 + 3 + 2/3 = 23/3

Total Resistance of the circuit = 7.67 ohm

Ques. In which circuit, the total resistance is greater than the largest resistance in the circuit? (2 Marks)

Ans. In series circuits, the total resistance is the sum of all the resistance in the circuit, hence the total is greater than the largest resistance.

Ques. In which circuit, the total resistance is smaller than the smallest resistance in the circuit? (2 Marks)

Ans. In a parallel circuit, the equivalent resistance is the sum of the reciprocals of all the resistances in the circuit. Hence it is smaller than the smallest resistance in the circuit.

Ques. Which is the most cost-efficient connection? (2 Marks)

Ans. Series connection is the most cost-efficient connection, due to the following reasons.

  • Less wiring is needed.
  • Less space is needed.

Ques. What are the advantages of parallel connections? (2 Marks)

Ans. Some of the advantages of parallel connections are:

  • If any one of the cells connected in parallel is damaged or disconnected, the other cells are not affected.
  • If the cells are connected in parallel they will not exhaust easily.
  • It has less internal resistance.

Ques. What are the disadvantages of parallel connections? (2 Marks)

Ans. Some of the disadvantages of parallel connections are:

  • The voltage developed will not be increased by increasing the number of cells in the parallel combination.
  • The output power is based on one cell. Therefore, the brightness of the bulb connected will not be very high.

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CBSE CLASS XII Related Questions

  • 1.
    The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

      • \( \frac{\alpha}{r} \)
      • \( \frac{\alpha r^2}{2} \)
      • \( \frac{\alpha}{2r^2} \)
      • \( -\frac{\alpha}{r} \)

    • 2.
      Draw a plot of frequency \( \nu \) of incident radiations as a function of stopping potential \( V_0 \) for a given photoemissive material. What information can be obtained from the value of the intercept on the stopping potential axis?


        • 3.
          Two point charges \( 5 \, \mu C \) and \( -1 \, \mu C \) are placed at points \( (-3 \, \text{cm}, 0, 0) \) and \( (3 \, \text{cm}, 0, 0) \), respectively. An external electric field \( \vec{E} = \frac{A}{r^2} \hat{r} \) where \( A = 3 \times 10^5 \, \text{V m} \) is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.


            • 4.
              In an intrinsic semiconductor, carrier’s concentration is \( 5 \times 10^8 \ \text{m}^{-3} \). On doping with impurity atoms, the hole concentration becomes \( 8 \times 10^{12} \ \text{m}^{-3} \).

              [(a)] Identify (i) the type of dopant and (ii) the extrinsic semiconductor so formed.

              [(b)] Calculate the electron concentration in the extrinsic semiconductor.


                • 5.
                  Two wires of the same material and the same radius have their lengths in the ratio 2:3. They are connected in parallel to a battery which supplies a current of 15 A. Find the current through the wires.


                    • 6.
                      Two cells of emf 10 V each, two resistors of 20 \( \Omega \) and 10 \( \Omega \), and a bulb B of 10 \( \Omega \) resistance are connected together as shown in the figure. Find the current that flows through the bulb.
                      Two cells of emf 10 V each, two resistors

                        CBSE CLASS XII Previous Year Papers

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