Combination of Cells in Series and in Parallel

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Jasmine Grover

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Cells are said to be connected in series when they are joined end to end so that the same quantity of electricity must flow through each cell. Whereas, cells are connected in parallel when they are joined positive to positive and negative to negative such that current is divided between the cells. There are methods to calculate the equivalent resistances when multiple resistances are connected in series or parallel or sometimes in a combination of series and parallel.

Read More: Difference between in physics


What are Cells?

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A cell is a device that uses chemical energy to generate electricity and maintains the flow of charge in a circuit. The energy is obtained from the chemical reactions that take place within these cells. Cells are commonly referred to as batteries, which are an essential day to day product required for a number of tasks. It is important to note that a battery is a collection of cells. And when placed in different circuits (in series or parallel), it functions differently. 
 

A cell consists of electrodes and electrolytes. The electrodes pass current in the circuit as they are conductors. The electrode which has higher potential is called anode or positive terminal of the cell. The electrode which has lower potential is called cathode or negative terminal of the cell.

A cell supplies energy when it is connected to an external load. The flow of current outside the cell is always from anode to cathode, while inside the cell the current flows from cathode to anode.

Discover about the Chapter video:

Current Electricity Detailed Video Explanation:

Also Read:


Electromotive Force or EMF

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EMF or electromotive force is defined as the potential difference between the positive and negative electrodes when there is no current in the cell or when the cell is in an open circuit. The flow of the current in the cell is initiated by the EMF of the cell.

The SI unit of EMF is Volts (V). The formula of EMF is- 

ε = E/Q

Where, ε = emf (Volts), E= Energy (Joules) and Q= charge (Coulombs).

The formula of EMF can also be written as,

ε = IR + Ir

Where,

ε = EMF (Volts),

I = Current (Ampere),

R = Load resistance in ohms,

r is the internal resistance of a cell in ohms.

The video below explains this:

Electromotive Force Detailed Video Explanation:


Internal Resistance

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Internal resistance is defined as the resistance offered by the electrolyte and electrodes when the current flows in the circuit. The flow of the current is opposed by the electrolytes and electrodes. The internal resistance is denoted by the letter ‘r’.

Let's consider a cell having two electrolytes which are connected to each other by an external resistance which is denoted by ‘R’. The current will flow from cathode to anode.

Closed Circuit: V = V1 + V2 – Ir 

Where (V1+V2) = initial potential difference,

Ir = potential drop across the internal resistance.

Therefore V = ε – Ir

Here ε=Emf, V=potential difference between 2 electrodes or it is the voltage drop across external resistance ‘R’.

⇒ IR = ε – Ir

Only when the circuit is closed the internal resistance will play its role. It varies from cell to cell.It is negligible where ε >> Ir.


EMF and Internal Resistance Equation

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The emf denoted as ε and the internal resistor with resistance r which is connected in series modifies the cell. The resistance R with an external load resistor is also connected across the circuit. The potential difference, V across the cell is represented as:

V = V+ + V – Ir

This represents the voltage drop present due to the internal resistance.

We know that 

ε = V+ + V = I(R + r)

ε = IR + Ir

= V + Ir

V = ε – Ir

Hence, V = ε – Ir

Here V is the potential difference across the circuit, ε is the emf, I is the current flowing through the circuit and r is internal resistance. 

In most instances, the internal resistance of a cell is not considered because ε >> Ir. Internal resistance usually changes in its value from cell to cell

Read More: Resistors in SeriesResistors in Parallel


Cells in Series and Parallel

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There are mainly two types of Connections of cells i.e. series and parallel. Cells can be connected in series, parallel or a combination of both. 

In a series circuit, electrons can travel only in one path. Here, the current passing through each resistor will be the same. However the voltage across resistors in a series connection will be different in each resistor. Thus circuits connected in series ensure that they do not overheat easily. 

In a parallel circuit electrons travel in several branches. In the case of a parallel circuit, the voltage remains the same across each resistor in the circuit and the cells can be arranged in different forms. Hence parallel circuits are used as a current divider and are easy to connect or disconnect a new cell or other component. This does not affect the other elements in the circuit. However, it uses a lot of wires and that is why the circuit becomes complex.

Cells in Series

When multiple cells are arranged in a manner where the positive terminal from the first cell is connected to the negative terminal of the next cell, and this sequence continues, it is called a series. Specifically, this arrangement is known as a series combination.

The arrangement is as shown below:

Let the EMF of the first cell = ε1 and EMF of the second cell = ε2.

Consider 3 points A, B and C as shown in the figure. The potential at point A= VA, potential at point B=VB and potential at point C=VC

Let us assume that the circuit is closed. As a result, there will be some flow of current. Resistance will be offered by the cell itself. 

Internal resistance of first cell =r1 and Internal resistance of second cell=r2.

Equation for potential difference ε1(first cell):-

For an open circuit

Potential difference between points A and B (V– VB) = potential difference between the 2 electrodes.

Therefore VA – VB =  ε1 – Ir1… Eq (i)

Where Ir1 = voltage drop across the internal resistance.

Similarly for equation for potential difference E2(second cell):-

VB – Vc =  ε2 – Ir...Eq (ii)

Adding (i) and (ii), 

VA – Vc = ε1 – Ir+ ε2 – Ir2

VA – Vc = ε1 + ε2 – I(r+ r2) ....Eq (iii)

Assuming if there are only 2 cells then the EMF = εequivalent= (ε1 + ε2) and resistance=requivalent= (r1+r2).

Using (iii)

VA - Vc = εequivalent–Irequivalent … Eq(iv)

By comparing RHS of equations (iii) and (iv), we get

εequivalent= (ε1 + ε2) and requivalent= (r1+r2)

For cells arranged in series

The sum of the individual EMFs is the equivalent EMF. The sum of the individual resistances gives the equivalent resistance.

Let n be the number of cells that are connected in series and connected across an external resistance R.

Internal resistance of each cell =r, Emf =ε

Total resistance = R +requivalent

Where R=external resistance of the external circuit.

Total resistance = R + nr where nr(sum of all the n resistances).

Thus the effective Emf of cells = nε, where nε (EMF of each cell).

Therefore current flowing through resistance

R(I) = \(\frac{Effective EMF}{Total Resistance} = \frac{nE}{R + nr}\)

Case 1: When R>>nr

Then I = \(\frac{nE}{R}\)  ignoring nr.

⇒ I = \(\frac{nE}{R}\)

Here (ε/R) = current caused by a single cell across external resistance.

So, the current becomes n times the current caused by the single cell.

Case 2: When 

R< I

I = \(\frac{n\varepsilon}{nR}\)

I = \(\frac{\varepsilon}{r}\)

Current present in the circuit is equal to the current due to a single cell across.

Conclusion

When cells are arranged in the series, the current flowing through the circuit depends upon the value of external and internal resistance. 

Cells in Parallel

A series of cells is said to be in parallel when all the positive terminals are connected together and all the negative terminals are connected together. 

In lieu of this arrangement, the potential difference across each of these cells will be the same.

Potential difference (first cell) =ε1; potential difference (second cell) =ε2.

Internal resistance (first cell) =r1; Internal resistance (second cell) =r2

The cell internal resistance is offered to the cell as current is flowing through it.

Equation for potential difference ε1(first cell):-

Potential difference between 2 points A and B.

Therefore V = ε1 – I1r1  (for first cell)

(Current I gets divided into 2 parts I1 and I2; I1 across ε1).

I1 = (ε– V)/r…..Eq (i)

Equation for potential difference ε2(second cell):

V = (ε2 – I2)r2 (current I2 flows across ε2)

I2 = (ε2 – V)/r…...Eq(ii)

Also, I = I1 + I2

Therefore using (i) and (ii)

I = (ε2 – I2)r+ (ε2 – V)/r2

⇒ ((ε1/r1) + (ε2/r2) – (V(1/r1) – (1/r2))

⇒ V = (I/(1/r1) + (1/r2)) + ((ε1/r1) + (ε2/r2) + (1/r2))

V = I(r1r2/r1 + r2) + (ε1r2 + ε2r1)/r1 + r2)…….Eq(iii)

Consider only 1 cell instead of 2 cells, i.e. 

εequivalent= (ε1 and ε2) and resistance = requivalent = (r1 and r2).

Therefore V= εequivalent - Irequivalent ...Eq (iv)

From Equation (iii),

V = I / ((1/r1) + (ε2/r1) + (ε2r2)) /  ((1/r1) + (1/r1))…….Eq (v)

After simplifying and Comparing (v) and (iv),

requivalent = (r1 + r2) / (r1 + r2), εequivalent = (ε1r1 + ε2r2)/r1 + r2)

Special Case

Let n be the number of cells connected in a parallel combination across an external resistance R. Let ε be the internal Emf of each cell and r be the internal resistance

Using εequivalent \(\frac{(\varepsilon_1r_1 + \varepsilon_2r_2)}{r_1 + r_2}\) 

Putting r1 = r2 = r, ε= ε= ε; 

εequivalent \(\frac{\varepsilon r}{n}\)

Total resistance=R + requivalent = \(\frac{r}{n}\) 

Current through R, I = (Effective Emf)/(total resistance) = (ε) / (R + (r/n)) = (nε) / (nR + r)

  • Case 1: R<<r; I = nε/r
  • Case 2: R>>>r; I = ε/r

Important Note:

  • If cells are arranged in series, it increases the voltage.
  • If cells are arranged in parallel, it increases the current.

The video below explains this:

Kirchhoff's Laws Detailed Video Explanation:

Also read:


Things to Remember

  • A cell is a device that uses chemical energy to generate electricity and maintains the flow of charge in a circuit.
  • Cells are said to be connected in series when they are joined end to end so that the same quantity of electricity must flow through each cell.
  • Cells are connected in parallel when they are joined positive to positive and negative to negative such that current is divided between the cells.
  • Internal resistance is defined as the resistance offered by the electrolyte and electrodes when the current flows in the circuit.
  • There are mainly two types of Connections of cells i.e. series and parallel. Cells can be connected in series, parallel or a combination of both. 

Previous Year Questions


Sample Questions

Ques. A cell has EMF 1.5 V. It can deliver a maximum current of 3 A. What will be the internal resistance of the cell?(2 marks)

Ans. For maximum current, the load resistance should be 0.

E = Ir

Rearranging the equation, we get

E=Ir

Substituting the values, we get

E=1.53=0.5Ω

Internal resistance of the cell = 0.5 Ω.

Ques. List the factors affecting the internal resistance of the cell.(2 marks)

Ans. The internal resistance of a cell depends on the following factors:

  • The surface area of the electrodes
  • The distance between the electrodes
  • Nature and the concentration of the electrolyte
  • The temperature of the electrolyte

Ques. A cell has an emf of 5.0 volt and an internal resistance of 1.0 volt. Its terminals are joined through a 9 Ω resistor. Calculate the potential difference across the terminals.(2 marks)

Ans. E/R+r= 5/9+1A= 1/2A

V=E-Ir= 5-1/2*1

V=4.5 volt

Ques. There are 27 resistors with internal resistance 0.4 Ω and an external resistance 1.2 Ω. What is the most effective way of grouping them?(2 marks)

Ans. Let n number of cells have m parallel branches.

mn= 27 (1)

mR=nr (2)

1.2m=n (0.4)

3m=n (3)

m=3 (derived from equation 1)

n= 9 (derived from equation 3)

Ques. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?(2 marks)

Ans. The advantages of parallel connection are: 

(a) In parallel circuit, if one electric appliance stops working due to some defect, then all other appliances keep working normally.

(b) In parallel circuits, each electrical appliance has its own switch due to which they can be turned on or off without affecting other appliances. 

(c) Each electrical appliance gets the same voltage (220 V) as that of the power supply line.

(d)While using parallel connection in electrical appliances, the overall resistance of the circuit is reduced which leads to a higher current from the power supply.

Ques. A cell of emf ‘E’ and internal resistance V is connected across a variable resistor ‘R’. Plot a graph showing variation of terminal voltage ‘V’ of the cell versus the current ‘I’. Using the plot, show how the emf of the cell and its internal resistance can be determined. (All India 2014, 2 marks)

Ans. Graph between ‘V’ and I

Intercept of Y-axis will give the value of ‘E’; while the slope of the line will give the value of internal resistance.

Ques. Two cells of emfs 1.5 V and 2.0 V having internal resistance 0.2 ? and 0.3 ? respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell. (Delhi 2016, 2 Marks)

Ans. Given that E1= 1.5V, E2= 2.0V 

r1= 0.2 Ω, r2= 0.3 Ω

Find Emf and Internal Resistance r when connected in parallel- 

E= E1r2+ E2r1/r1+r2

= 1.5* 0.3+2*0.2/0.2+0.3

=0.85/0.5

= 10.7V

r= r1r2/r1+r2

 =0.2* 0.3/0.2+0.3

= 0.12Ω

Ques. A number of identical cells n, each of emf e, internal resistance r connected in series are charged by a d.c. source of emf elr using a resistor R.
(i) Draw the circuit arrangement.
(ii) Deduce the expressions for
(a) the charging current and
(b) the potential difference across the combination of the cells. (Delhi 2008, 3 marks)

Ans.

Draw the circuit arrangement. (ii) Deduce the expressions for (a) the charging current and (b) the potential difference across the combination of the cells.

Ques. A cell of emf E and internal resistance r is connected across a variable resistor Plot a graph showing variation of terminal voltage V of the cell versus the current I. Using the plot, show the emf of the cell and its internal resistance can be determined.[All India 2014, 2 marks]

Ans. 

cells in series and parallel

Ques. Two cells of emf 2E and E and internal resistances 2r and r respectively, are connected in parallel. Obtain the expressions for the equivalent emf and the internal resistance of the combination.[All India 2010 C, 3 Marks]

Ans.  For parallel connection total current

I= I1 + I2

Or,Eeq/req=E1/r1+E2/r2

Or,Eeq=reqE1/r1+reqE2/r2

=(r2E1+r1E2)/(r1+r2)

=4E/3

Equivalent internal resistance is simply

req=r1r2/(r1+r2)

=2r/3

Ques. A cell of emf E and internal resistance r draws a current Write the relation between terminal voltage V in terms of E, I and r.[Delhi 2013, 1 Mark]

Ans.  When a current I is Drawn from a cell of emf E and internal resistance r, then the relation between terminal voltage V, E, I and r will be 

V = ε – Ir

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CBSE CLASS XII Related Questions

1.

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

      2.
      Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the to charges is the electric potential zero? Take the potential at infinity to be zero.

          3.

          Three capacitors each of capacitance 9 pF are connected in series. 

          (a) What is the total capacitance of the combination? 

          (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

              4.

              A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s−1.

              1. What is the rms value of the conduction current?
              2. Is the conduction current equal to the displacement current?
              3. Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
              A parallel plate capacitor made of circular plates

                  5.

                  A tank is filled with water to a height of 12.5cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

                      6.
                      (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. 
                      (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

                          CBSE CLASS XII Previous Year Papers

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