Cells, EMF and Internal Resistance: Equations & Examples

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Cells, EMF, and Internal Resistance are the components that make up a circuit and help in the flow of electricity within the electric circuit. These three components are interrelated to each other in a circuit. Batteries or cells comprise internal resistance and potential difference (voltage).

  • Electric Cell is a device that can generate electrical energy from chemical processes or vice versa. 
  • EMF stands for ‘electromotive force’ which is the transfer of energy to a circuit per unit of electric charge.
  • Internal Resistance is the opposition to the current flow by the cells and batteries themselves leading to heat generation.

Read More: NCERT Solutions For Class 12 Physics Current Electricity

Key Terms: Cells, EMF, Internal Resistance, Electrical Circuit, Electricity, Electrochemical Cells, Voltage, Battery, Anode, Cathode


What are Cells?

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Electric Cell is also referred to as an “Electrochemical Cell” or “Electric Power Supply”. Cells are used to generate electricity as well as derive chemical reactions. One or more electrochemical cells are called batteries. Every cell has two terminals which are:

  • Anode: Anode is the terminal that forms an incoming channel for the current to enter the circuit.
  • Cathode: It is the terminal from where the electric current flows out, i.e. it provides an outgoing current flow from the circuit.

Cell and Battery

There are different types of cells and some of them are:

  • Electric Cells
  • Fuel Cells
  • Secondary Cells
  • Galvanic Cells
  • Photovoilatatie Cells
  • Solar Cells
  • Storage Cells
  • Primary Cells

Discover about the Chapter video:

Current Electricity Detailed Video Explanation:

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Primary Cells and Secondary Cells

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Primary Cells and Secondary Cells are the two types of electrochemical cells. The key difference between a primary cell and the secondary cell is that primary cells cannot be charged again but secondary cells are the ones that are rechargeable.

Primary Cell

  • Primary cells cannot be charged again, so they can only be used once.
  • The chemical reaction will be irreversible and the active materials may not return to their original form in the primary cell.
  • Dry cells, alkaline cells, and mercury cells are different examples of primary cells.
  • Primary cells cost low and can be used easily.
  • However, they are not suitable for heavy loads.

Primary Cell

Read More: ​Difference Between Primary Cells & Secondary Cells

Secondary Cell

  • Secondary cells can be recharged and reused again and again.
  • The chemical reaction of secondary cells is reversible.
  • Examples of secondary cells are lead-acid cells and fuel cells.
  • Lead acid cells are widely used in vehicles and other applications requiring a high load current value.
  • Car batteries and power supply stands are secondary cells.
  • Solar cells are secondary cells that convert the sun’s light energy into electrical energy.
  • Though the cost of the secondary cells is higher compared to primary cells, they can be used for a long period of time.
  • The usage of the secondary cells is more complex.

Secondary Cell

Read More: Current Electricity 


Combinations of Cell

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In order to produce the desired value of voltage and current, two or more voltage sources are used in different combinations. Cells can be connected in two basic types of combinations which are

  1. Series Combination
  2. Parallel Combination

Series Combination of Cells

  • A higher resulting voltage is produced for the cells connected in series.
  • The cells that are damaged can be easily identified and can therefore be easily replaced as they break the circuit.
  • If any of the cells in the circuit is damaged, the entire connection may be affected.
  • The cells which are connected in series get easily exhausted and so they do not last long.
  • This type of combination is not used in house wiring.

Series Combination of Cells

Parallel Combination of Cells

  • If the cells are connected in parallel and any one of the cells is damaged in the circuit, it will not affect the whole connection.
  • The cells which are connected in parallel do not exhaust easily and thus they last longer.
  • The voltage developed by the cells in parallel connection cannot be increased by increasing the number of cells present in the circuit.
  • It is because they do not have the same circular path.
  • In a parallel connection, the connection provides power based on one cell. So the brightness of the electronic bulb will not be high.

Parallel Combination of Cells


What is EMF?

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EMF or Electromotive force is defined as the potential difference between the two terminals of the battery in an open circuit.

  • Anode has a positive potential (V+) and cathode has a negative potential (V-). 
  • It is the potential difference between the anode and the cathode when there is no current flowing through it.
  • It measures the energy which is transferred to the charge carried in the cell or a battery.
  • It is the energy in joules divided by the charge in coulombs.
  • It acts as the initiating force for the current to flow.
  • EMF is denoted by the symbol ‘ε’.

EMF Formula is given as

ε = E/Q

Where

  • ε: Electromotive Force
  • E: Energy
  • Q: Charge

EMF

The video below explains this:

Electromotive Force Detailed Video Explanation:

Read More: Current Electricity Important Questions 


What is Internal Resistance?

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Internal Resistance is the resistance present within a battery that resists the current flow when connected to a circuit.

  • It causes a voltage drop when current flows through it.
  • It is the resistance provided by the electrolyte and electrodes which is present in a cell.
  • It is denoted by the letter ‘r’.

Internal Resistance

Internal Resistance Formula is given as

Internal Resistance (r) = (E – V)/I

Where

  • E is the emf of the device
  • V is the potential difference between the device.
  • I is the current in the device.

Internal Resistance is the result of the resistance in the battery or the accumulation in the battery. The equation used to derive this is as follows:

V = (E – Ir)

Internal Resistance and EMF

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The relation between internal resistance and EMF can be explained with the help of two cases given as follows:

Case 1: When Cell acts as a Source

In this case, there will be resistance to the flow of current due to the material of the cell. It is referred to as internal resistance which is measured in ohms.

Assume that the cell acting as the source of current is connected to a closed circuit. When the cell gives out current (discharging) due to the internal resistance, the measured potential difference across terminals A and B will be less than the actual EMF of the cell.

VAB = ε – ir

Where

  • VAB: Terminal Voltage or Potential Difference across Terminals A and B.
  • ε: EMF of the Cell.
  • r: Internal Resistance of the Cell.
  • i: Current through the Circuit
  • ir: Voltage Drop due to Internal Resistance

In the above expression, the negative sign is used as the current flows from the positive terminal A to the negative terminal B. Thus, it can be understood that the terminal voltage is less than the EMF of the cell when the current flows out of the positive terminal of the cell.

Read More: Important MCQs on Current Electricity

Case 2: When Cell acts as the Load

A cell acts as a load when it is charging. In such a situation, the current enters the positive terminal of the cell.

Thus, the terminal voltage will be calculated as

VAB = ε + ir

Where

  • VAB: Terminal Voltage
  • ε: EMF of the Cell
  • r: Internal Resistance of the Cell
  • i: Current through the Circuit

When the current enters the positive terminal of the cell, the terminal voltage is greater than the EMF of the cell.

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Solved Examples on Cells, EMF and Internal Resistance

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Given below are a few solved examples on Cells, EMF, and Internal Resistance to get a better idea of the concept:

Example 1: What condition does the terminal voltage of a cell in an open circuit follow?

  1. Less than the emf
  2. More than the emf
  3. Depends on internal resistance
  4. Equal to the emf

Solution: (d) Equal to the emf.

Explanation: In the condition of the terminal voltage of a cell in an open circuit, it can be said that the terminal voltage will be equal to the emf of the cell. This is because the circuit is open, thus it comes with no drop across the internal resistance.

Example 2. A student performs an experiment by plotting the V-I graph of three nichrome wire samples that have resistances R1, R2, and R3 respectively. Which one of the following is true?

V-I graph of three nichrome wire samples

  1. R= R2 = R3
  2. R> R2 > R1
  3. R> R1 > R3

Solution: (b) R> R2 > R1.

Explanation: Rslope is maximum, while R3 is minimum.

Thus, (Slope)R1 >(Slope)R2 > (Slope)R3

We are aware that, Resistance of wire = \( {1 \over Slope\,of\, the\, I-V \,Graph}\)

Thus, R> R2 > R1.

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Things to Remember

  • Electric cell is a device used to convert chemical energy into electrical energy. 
  • A battery is a combination of one or more electrochemical cells.
  • EMF or Electromotive force is the potential difference between the two terminals of the battery in an open circuit.
  • Electromotive force is given as ε = E/Q.
  • Internal Resistance is a resistance present within the battery that resists the current flow when connected to a circuit.
  • Internal resistance is expressed as Internal Resistance (r) = (E – V)/I.

Previous Year Questions 

  1. In a potentiometer, the null point is received at the 7th wire… (DUET 2007)
  2. The electrical permittivity and magnetic permeability of free space are... (DUET 2003)
  3. In the circuit given in figures, 1 and 2 are ammeters. Just after key K… (KEAM 1999)
  4. Three resistances each of 4Ω are connected to form a triangle… (NEET 1993)
  5. A potentiometer consists of a wire of length 4 m and resistance… (NEET 1999)
  6. A battery of e.m.f 10 V and internal resistance 0.5Ω is connected... (NEET 1992)
  7. In the circuit shown below, the current in the 1Ω resistor is… (NEET 1988)
  8. In the network shown in the figure, each of the resistance is equal to… (NEET 1995)
  9. Kirchhoff's first and second laws of electrical circuits are consequences of… (NEET 2006)
  10. Current through 3Ω resistor is 0.8 ampere, then potential drop... (NEET 1993)

Sample Questions

Ques. The potential difference across a cell is 1.8 V when a current of 0.5 A is drawn from it. The potential difference falls to 1.6 V volt when a current of 1.0 A is drawn. Find the EMF and internal resistance of the cell. (3 Marks)

Ans. Given that,

  • i = 0.5 A
  • V = 1.8

Suppose the EMF of the cell is ε and internal resistance is r. Now we know,

V = ε – ir

1.8 = ε – 0.5r ….(1)

Now when,

  • i = 1
  • V = 1.6

1.6 = ε – r …..(2)

Solving equations (1) and (2)

E = 2 Volt and r = 0.4 ohm

Thus, the EMF is 2 Volts and the internal resistance is 0.4 ohm.

Ques. A common dry cell produces a voltage of:
(a) 1.5 V
(b) 30 V
(c) 60 V
(d) 3 V (2 Marks)

Ans. (a) 1.5 V

Explanation: A common dry cell is a type of electrochemical cell that is commonly used for many home appliances and portable devices. They are often used in form of batteries. As per standards, a common dry cell has a constant voltage of 1.5

Ques. What is the resistance of a 1000 W and 120 V toaster? (3 Marks)

Ans. Given that

  • P = 1000 W
  • V = 120 V

Power is given by the formula, 

P = VI

1000 = 120 x I

I = 8.333 A

Now from Ohm’s law,

V = IR

R = V/I

R = 14.4 ohm

Thus, the resistance of a 1000 W and 120 V toaster is 14.4 ohm.

Ques. State the difference between potential difference and electromotive force. (2 Marks)

Ans. An electromotive force, popularly known as EMF, is defined as the total voltage in an electric circuit produced by the battery or source. On the other hand, Potential difference is defined as the work done between two specific points in the circuit for moving a unit charge against the electric field. EMF is independent of the circuit's internal resistance while the potential difference is proportional to the circuit's resistance.

Ques. The potential difference across the cell when no current flows through the circuit is 3 V. When the current I = 0.37 Ampere is flowing, the terminal potential difference falls to 2.8 Volts. Determine the internal resistance (r) of the cell. (3 Marks)

Ans. We know that, 

e = V + Ir

e – V = Ir

(e – V)/I = r

Therefore, r = (3.0 – 2.8)/0.37 = 0.54 Ohm.

Due to the Internal Resistance of the cell, the electrons moving through the cell turns some of the electrical energy to heat energy. Therefore, the potential difference available to the rest of the circuit is:

V = E (EMF of Cell) – Ir (the p.d. across the internal resistor)

Ques. Name the two types of electric cells. (2 Marks)

Ans. Primary Cells and Secondary Cells are the two types of electric cells.

  • Primary Cell: A primary cell cannot be charged again and again. Some examples of primary cells are Voltaic cells, Daniel cells, and Leclanche cells.
  • Secondary Cell: These cells can be charged again and again. Some examples of secondary cells are Alkali and acid accumulators.

Ques. Two electric bulbs P and Q have their resistances in the ratio of 1:2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs. (2 Marks)

Ans. Let the bulbs P and Q have resistance be R and 2R respectively. The current passing through each and every bulb will be equal as they are joined in series. 

Let us take the current as I:

P1/P2 = I2R1/ I2R2 = R1/R2 = 1/2

P1:P2 = 1:2

Ques. Define terminal voltage. (2 Marks)

Ans. Terminal voltage is defined as the potential difference that is measured across the terminals of the battery. When no current is drawn from the battery, the terminal voltage is equal to the EMF value of the battery. However, when a current is drawn, then the resultant potential difference will appear across terminal voltage as the difference between the EMF and the potential difference across the internal resistance due to current flow.

Ques. Two cells of EMFs 1.5 V and 2.0 V having internal resistances 0.2 Ω, and 0.3 Ω, respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell. (3 Marks)

Ans. We have

  • E1 =1.5V E2 = 2.0V
  • r= 0.2Ω r2= 0.3Ω

Equivalent resistance of parallel combination: 1/req = 1/r1 + 1/ r2

Therefore 1/req = 1/0.2 + 1/0.3 = (0.2+0.3)/(0.2*0.3)

req = 0.12Ω

Equivalent Emf: Eeq = (E/r + E/r)req

Therefore Eeq = [(1.5/0.2) + (2.0/0.3)] x 0.12

Eeq = (7.5+6.67) * 0.12 = 1.7 V

Ques. What is Internal Resistance? List the factors affecting internal resistance? (2 Marks)

Ans. Internal resistance is defined as the resistance offered by the cell due to the nature of the material of the cell. It is affected by the given factors:

  • Distance between electrodes
  • Nature of electrode and electrolyte
  • Area of electrode

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CBSE CLASS XII Related Questions

1.
(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.)

      2.

      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?

          3.

          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?

              4.
              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.

                  5.
                  A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C distributed uniformly on its surface. What is the electric field ?
                  1. inside the sphere
                  2. just outside the sphere
                  3. at a point 18 cm from the centre of the sphere?

                      6.

                      An object of size 3.0 cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

                          CBSE CLASS XII Previous Year Papers

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