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Physics Topics are also essential for space exploration, allowing scientists to study phenomena such as gravitational waves and cosmic rays.
What is the Relationship Between Lost Volts and Internal Resistance? What Do You Mean by Internal Resistance of a cell?
The figure shows an arrangement in which a resistance R is joined between points A and B of a conducting wire which terminates at the two poles of a cell of emf E [Fig.]. The resistance of the connecting wire is negligible.
A current I passes through R and completing the circuit from the negative pole to the positive pole inside the cell. When the current flows through the cell the elements of the cell oppose it to some extent indicating that inside the cell a resistance is developed. This is the internal resistance of the cell (r).
The equivalent resistance of the circuit = external resistance + internal resistance of the cell = R + r.
Current, I = \(\frac{E}{R+r}\) or, E = IR + Ir
Now, potential difference of the external circuit i.e., the potential difference across the resistance, R is V = IR .
∴ E = V + Ir
or, V = E – Ir ……….. (1)
V is the potential difference of the cell. Obviously, the value of V is less than E. It indicates that the whole emf of the cell is not obtained as potential difference in the external circuit. A poten-tial of magnitude Ir is lost inside the cell for the internal resistance (r). This is called internal potential drop or lost volt of the cell. This is so called because this portion of the emf of the cell does not contribute to the current through the external resistance.
Discussion:
Connection of voltmeter: If the two ends of a voltmeter are connected to the two poles of a cell, the reading it indicates is not the value of E (emf of the cell) but the value of V (potential difference of the cell) because the voltmeter acts here as the external circuit.
Negligible internal resistance: If the internal resistance of the cell is very small in comparison to the external resistance then r may be taken as zero. Then Ir = 0 and no potential is lost inside the cell. So, emf of the cell and potential difference of the cell are equal. For example—In a lead-acid accumulator the magnitude of the internal resistance ranges from 0.01 Ω to 0.1 Ω. So, this resistance may be neglected in most of the electrical circuits.
Definition of electromotive force: in an open circuit, no current flows in the external circuit i.e., Ir = 0. Then lost volt, Ir – 0 and V = E. From this relation emf of the cell may be defined as follows.
The potential difference between the two poles of a cell in an open circuit (when it does not deliver current in an external cir-cuit) is called the emf of the cell.
Limit of Current: In the equation V = E – Ir if the magnitude of I or r is very high then the magnitude of V becomes very small. If Ir becomes equal to E, then V = 0 . Thus the cell cannot produce any potential difference in the external circuit. So, for every cell I has a maximum limit. For example, the emf of a dry cell is 1.5 V and if its internal resistance is 3 fl, the maximum current it delivers in the external circuit is I = \(\frac{E}{r}\) = \(\frac{1.5}{3}\) = 0.5 A. Hence primary cells are not used where high currents are required.
Voltage source and current source: Consider a battery of emf E with an internal resistance r [Fig.], to which a vari-able resistor (R) is connected. The current in the circuit will be,
I = \(\frac{E}{R+r}\) …. (1)
The potential differences between the two poles of the battery i.e., the termi-nal potential differences of the resistance R is,
V = IR = \(\frac{E R}{R+r}\) ….. (2)
i) When r\(\ll\)R; we get from equation (2),
V = \(\frac{E}{1+\frac{r}{R}}\) ≈ constant
Hence, whatever be the value of R, we get a constant potential difference (or voltage) from the battery. Under this condition, the battery is taken as a voltage source.
ii) When r\(\gg\)R, we get from equation (1),
I = \(\frac{E}{r\left(1+\frac{R}{r}\right)}\) ≈ \(\frac{E}{r}\) = constant
Therefore, whatever be the value of R, we get a constant current in the external circuit. Under this condition, the battery acts as a current source.
Numerical Examples
Example 1.
A resistor is fabricated by connecting two wires of the same material. The radii of the two wires are 1 mm and 3 mm respectively and their lengths are 3 cm and 5 cm respectively. If the two ends of the resistor are connected to the two terminals of a battery of emf 16 V and of negligible internal resistance, what will be the potential difference between the two ends of the wire of shorter length?
Solution:
Resistance of wire, R = \(\rho \frac{l}{A}\) = ρ ᐧ \(\frac{l}{\pi r^2}\)
So, for the two wires of same material,
The two wires are connected in series. So, the ratio of the poten-tial difference across their ends is
In the circuit there are only two resistances. So the electromotive force,
E = V1 + V1 = 16 ∴ V1 = \(\frac{27}{32}\) × 16 = 13.5 V
So, the potential difference across the resistance of length 3 cm
= V1 = 13.5 V
Example 2.
A heater of resistance 140 Ω capable of carrying a cur-rent of 1.2 A is put in a dc mains of 210 V. Find out the minimum value of an additional resistance to be added to run the heater.
Solution:
Let the value of the minimum additional resistance to be added be x Ω.
We know, V = IR
or, 210 = 1.2(140 + x) [Here R = 140 + x]
or, 140 + x = \(\frac{210}{1.2}\) or, x = 175 – 140 or, x = 35 Ω
So, a minimum additional resistance of 35 Ω is to be added to run the heater.
Example 3.
To the parallel combination of two resistances 3 Ω and 1 Ω, a series combination of resistances 2.15 Ω and 1 Ω and a battery are connected. The internal resistance of the battery is 0.1 Ω and the emf is 2 V. Determine the values of currents flowing through the resistances. Draw the diagram of the circuit.
Solution:
Refer to Fig. for the circuit diagram. The equiva-lent resistance of the parallel combination of 3 Ω and 1 Ω is
This equivalent resistance is in series with the other resistances. So, the main current of the circuit is
I = \(\frac{2}{0.75+2.15+1+0.1}\) = \(\frac{2}{4}\) = 0.5 A
This main current flows through the resistances 2.15 Ω and 1 Ω connected in series.
Now, current through 3 Ω,
I1 = I × \(\frac{1}{3+1}\) = 0.5 × \(\frac{1}{4}\) = 0.125
and current through 1 Ω,
I2 = I – I1 = 0.5 – 0.125 = 0.375 A
Example 4.
The electromotive force of a cell is 2 V. The potential difference becomes 1.5 V when a resistance of 15 ft is added to the two ends of the cell. Determine the inter-nal resistance of the cell and the lost volt.
Solution:
Lost volt = E – V = 2 – 1.5 = 0.5 V
Current of the external circuit
\(=\frac{\text { potential difference of the external circuit }}{\text { resistance of the external circuit }}\)
= \(\frac{1.5}{15}\) = 0.1 A
This is current flowing through the cell.
So, internal resistance of the cell,
r = \(\frac{\text { lost volt (Ir) }}{\text { current }(I)}\) = \(\frac{0.5}{0.1}\) = 5Ω
Example 5.
In the given circuit diagram [Fig.], what is the current sent by the battery? [AIEEE 04]
Solution:
The equivalent circuit is shown in the Fig..
The equivalent resistance of the middle branch
= 1.5 + \(\frac{2 \times 6}{2+6}\) = 3Ω
With this 3Ω resistance, the outer 3Ω resistance on the right side is in parallel combination. So the equivalent resistance of the circuit
= \(\frac{3 \times 3}{3+3}\) = \(\frac{3}{2}\)Ω
∴ Main current of the circuit,
I = \(\frac{6}{\frac{3}{2}}\) = 4A
Comparison of Electromotive Force and Potential Difference
i) In a source of electricity, chemical or some other form of energy is converted into electrical energy. This produces the electromotive force. Again when a source of electricity is connected to an external circuit, then a potential difference is developed at the two ends of the circuit. As a result, the electrical energy of the source is converted into another form of energy in the external circuit.
ii) Due to the existence of electromotive force in the source of electricity, potential difference develops in the external cir-cuit i.e., if the electromotive force is called the ’cause] the potential difference may be called the ‘effect’
iii) The unit of electromotive force and potential difference is the same and this unit is volt.
iv) An amount of potential is lost inside the source of electricity due to its internal resistance. So, the entire electromotive force cannot be obtained as the potential difference in the external circuit i.e., potential difference can never be greater than the electromotive force.
v) When the source of electricity does not send current in the external circuit i.e., when the circuit is open, then no potential is lost due to the internal resistance of the source. In that case the potential difference between the two ends of the source becomes equal to its electromotive force.
vi) The electromotive force of an electric cell without any defect is a constant quantity. But from the equation V = E – Ir it can be said that if the current flowing through the circuit increases or decreases, the potential difference at the two ends of the circuit decreases or increases accordingly.