CBSE NCERT NOTES CLASS 12 PHYSICS CHAPTER 3
CURRENT ELECTRICITY
Electric current
The rate of flow of charge in any conductor is called electric current.
If Q is the net amount of charge flowing across the area in the forward direction in the time interval t, then,
Current, I = $\frac{\mathrm{Q}}{\mathrm{t}}$
Instantaneous current
$$\mathrm{I}=\mathrm{}\underset{\mathrm{\Delta}\mathrm{t}\mathrm{\to}0}{\mathit{lim}}\frac{\mathrm{\Delta}\mathrm{Q}}{\mathrm{\Delta}\mathrm{t}}$$
I is defined to be the current across the area in the forward direction. (If it turns out to be a negative number, it implies a current in the backward direction.)
If a charge q revolves in a circle with frequency ν, the equivalent current, then i = qν
SI unit of current is Ampere (A).
OHM’S LAW
This law states that the current passing through a conductor is directly proportional to the potential difference cross its ends, provided the physical conditions like temperature, remain unchanged.
Imagine a conductor through which a current I is flowing and let V be the potential difference between the ends of the conductor. Then,
V ∝ I
Or V = IR
Or $\frac{\mathrm{V}}{\mathrm{I}}$ = constant = R
$$\mathrm{O}\mathrm{r}\mathrm{}\mathrm{I}\mathrm{}=\frac{\mathrm{V}}{\mathrm{R}}$$
where the constant of proportionality R is called the resistance of the conductor.
The SI unit of resistance is ohm, and is denoted by the symbol Ω.
Resistance
It is a property of a conductor by virtue of which it opposes the flow of current through it. It is equal to the ratio of the potential difference applied across its ends and the current flowing through it.
$$\mathrm{R}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}=\mathrm{}\frac{\mathrm{P}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}}{\mathrm{C}\mathrm{u}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}}$$
Or R = $\frac{\mathrm{V}}{\mathrm{I}}$
Ohm
It is the S.I. unit of resistance. A conductor has a resistance of one ohm if a current of one ampere flows through it on applying a potential difference of one volt across its ends.
1 ohm = $\frac{1\mathrm{}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{t}}{1\mathrm{}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{e}}$
⇒ 1Ω = $\frac{1\mathrm{V}}{1\mathrm{A}}$
Factors affecting the resistance of a conductor
(A) At a given temperature, the resistance R of a conductor depends
i) Directly on its length l i.e.
$$\mathrm{R}\mathrm{}\propto \mathrm{}\mathrm{l}$$
ii) Inversely on its area of cross-section A i.e.
$$\mathrm{R}\mathrm{}\propto \frac{1}{\mathrm{A}}$$
iii) On the nature of material of the conductor on. On combining the above factors, we get
$$\mathrm{R}\mathrm{}\propto \frac{\mathrm{l}}{\mathrm{A}}$$
Therefore, we can write,
$$\mathrm{R}\mathrm{}=\mathrm{}\mathrm{\rho}\mathrm{}\times \frac{\mathrm{l}}{\mathrm{A}}$$
The proportionality constant ρ (Rho) is called resistivity of conductor.
(B) Temperature of the conductor.
Resistivity
It is defined as the resistance offered by a cube of a material of side 1 m when current flow perpendicular to its opposite faces. It is a characteristic property of the material. Its S.I. unit is ohm-meter (Ωm).
$$\mathrm{Resistivity,\; \rho \; =}\frac{\mathrm{R}\mathrm{}\mathrm{A}}{\mathrm{L}}$$
The metals and alloys have very low resistivity in the range of 10^{–8} Ωm to 10^{–6} Ωm. They are good conductors of electricity.
Insulators like rubber and glass have resistivity of above 10^{12} Ωm.
The substances with resistivity from 10^{-6} Ωm to 10^{12} Ωm are called semi-conductors.
- Both the resistance and resistivity of a material vary with temperature.
- Resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures. Hence they are used in electrical heating devices, like electric iron, toasters etc.
- Tungsten is used almost exclusively for filaments of electric bulbs.
- Copper and aluminium are used for electrical transmission lines, due to their low resistivity and reasonable cost.
Current density
Current per unit area (taken normal to the current), $\frac{\mathrm{I}}{\mathrm{A}}$, is called current density and is denoted by $\overrightarrow{\mathrm{j}}$.
The SI units of the current density are A/m^{2}. It is a vector quantity.
If E is the magnitude of uniform electric field in the conductor whose length is l, then the potential difference V across its ends is E l.
E l = j ρ l
Or E = j ρ
Or vectorially,
$$\overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{j}}\mathrm{\rho}$$
$$\mathrm{\Rightarrow}\overrightarrow{\mathrm{j}}=\mathrm{\sigma}\overrightarrow{\mathrm{E}},$$
where σ ≡ $\frac{1}{\mathrm{\rho}}$ is called the conductivity.
Drift of electrons - origin of resistivity
In absence of Electric field, electrons in a conductor will be moving due to thermal motion during which they collide with the fixed ions. An electron colliding with an ion emerges with the same speed as before the collision. However, the direction of its velocity after the collision is completely random.
Thus, if there are N electrons and the velocity of the ith electron (i = 1, 2, 3, ... N ) at a given time is vi, then
$$\frac{1}{\mathrm{N}}\sum _{\mathrm{i}=1}^{\mathrm{N}}{\mathrm{V}}_{\mathrm{i}}=0$$
At a given time, there is no preferential direction for the velocities of the electrons. Thus on the average, the number of electrons travelling in any direction will be equal to the number of electrons travelling in the opposite direction. So, there will be no net electric current.
If an electric field is applied, the electrons will be accelerated due to this field by
$$\mathrm{a}=-\frac{\mathrm{e}\mathrm{E}}{\mathrm{m}}$$
Where –e is the charge and m is the mass of an electron.
Consider now the ith electron at a given time t. This electron would have had its last collision some time before t, and let ti be the time elapsed after its last collision.
If v_{i} was its velocity immediately after the last collision, then its velocity V_{i} at time t is
$${\mathrm{V}}_{\mathrm{i}}={\mathrm{v}}_{\mathrm{i}}-\frac{\mathrm{e}\mathrm{E}}{\mathrm{m}}{\mathrm{t}}_{\mathrm{i}}$$
The average of v_{i}’ s is zero since immediately after any collision, the direction of the velocity of an electron is completely random.
The collisions of the electrons do not occur at regular intervals but at random times. Let τ be the average time between successive collisions. Then at a given time, some of the electrons would have spent time more than τ and some less than τ. That is, the time t_{i} will be less than τ for some and more than τ for others as we go through the values of i = 1, 2 ..... N. The average value of t_{i} then is τ (known as relaxation time).
Drift velocity
Under the influence of electric field, the electrons move with an average velocity which is independent of time, although electrons are accelerated. This velocity is called the drift velocity. It is represented by v_{d}.
If v_{d} is average velocity then,
$${\mathrm{v}}_{\mathrm{d}}={\left({\mathrm{V}}_{\mathrm{i}}\right)}_{\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}\mathrm{}}$$
$$={{(\mathrm{v}}_{\mathrm{i}})}_{\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}}-\frac{\mathrm{e}\mathrm{E}}{\mathrm{m}}{\left({\mathrm{t}}_{\mathrm{i}}\right)}_{\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}}$$
$$=0-\mathrm{}\frac{\mathrm{e}\mathrm{E}}{\mathrm{m}}\mathrm{\tau}\mathrm{}$$
$$=-\mathrm{}\frac{\mathrm{e}\mathrm{E}}{\mathrm{m}}\mathrm{\tau}$$
There will be net transport of charges across any area perpendicular to E. Consider a planar area A, located inside the conductor such that the normal to the area is parallel to E.
In an infinitesimal amount of time $\mathrm{\Delta}$t, all electrons to the left of the area at distances upto |v_{d}|$\mathrm{\Delta}$t would have crossed the area.
If n is the number of free electrons per unit volume in the metal, then there are n $\mathrm{\Delta}$t |v_{d}|A such electrons.
The total charge transported across this area A to the right in time $\mathrm{\Delta}$t is –ne A |v_{d}| $\mathrm{\Delta}$t.
The amount of charge crossing the area A in time $\mathrm{\Delta}$t is by definition = I $\mathrm{\Delta}$t, where I is the magnitude of the current.
Therefore, we can write,
$$\mathrm{I}\mathrm{\Delta}\mathrm{t}\mathrm{}=\mathrm{}\u2013\mathrm{n}\mathrm{e}\mathrm{}\mathrm{A}\mathrm{}\left|{\mathrm{v}}_{\mathrm{d}}\right|\mathrm{}\mathrm{\Delta}\mathrm{t}\mathrm{}$$
Now, putting the value of v_{d}
$$\mathrm{I}\mathrm{\Delta}\mathrm{t}=\frac{{\mathrm{e}}^{2}\mathrm{A}}{\mathrm{m}}\mathrm{}\mathrm{\tau}\mathrm{}\mathrm{n}\mathrm{}\mathrm{\Delta}\mathrm{t}\mathrm{}\left|\overrightarrow{\mathrm{E}}\right|$$
$$\Rightarrow \mathrm{}\mathrm{I}=\frac{\mathrm{n}{\mathrm{e}}^{2}\mathrm{A}}{\mathrm{m}}\mathrm{}\mathrm{\tau}\mathrm{}\left|\overrightarrow{\mathrm{E}}\right|$$
Or
$$\left|\overrightarrow{\mathrm{j}}\right|=\frac{\mathrm{I}}{\mathrm{A}}=\frac{\mathrm{n}{\mathrm{e}}^{2}}{\mathrm{m}}\mathrm{}\mathrm{\tau}\mathrm{}\left|\overrightarrow{\mathrm{E}}\right|\mathrm{}$$
The vector $\overrightarrow{\mathrm{j}}$ is parallel to $\overrightarrow{\mathrm{A}}$ or $\overrightarrow{\mathrm{E}}$ and hence we can write
$$\overrightarrow{\mathrm{j}}=\frac{\mathrm{I}}{\overrightarrow{\mathrm{A}}}=\frac{\mathrm{n}{\mathrm{e}}^{2}}{\mathrm{m}}\mathrm{}\mathrm{\tau}\overrightarrow{\mathrm{}\mathrm{E}}\mathrm{}$$
Comparing with Ohm’s law (j = σ E), we get,
$$\mathrm{\sigma}=\frac{\mathrm{n}{\mathrm{e}}^{2}}{\mathrm{m}}\mathrm{}\mathrm{\tau}$$
Mobility of electrons
The drift velocity of electron per unit electric field applied is called mobility of electrons. It is represented by μ.
Mobility of electron
$$\mathrm{\mu}\mathrm{}=\frac{\left|{\mathrm{v}}_{\mathrm{d}}\right|}{\mathrm{E}}=\mathrm{}\frac{\mathrm{e}\mathrm{}\mathrm{\tau}}{\mathrm{m}}$$
Putting the value,
$$\left|{\mathrm{v}}_{\mathrm{d}}\right|=\frac{\mathrm{e}\mathrm{}\mathrm{E}\mathrm{}\mathrm{\tau}}{\mathrm{m}}\mathrm{}$$
Wet get,
$$\mathrm{}\mathrm{\mu}\mathrm{}=\frac{\frac{\mathrm{e}\mathrm{}\mathrm{E}\mathrm{}\mathrm{\tau}}{\mathrm{m}}}{\mathrm{E}}=\mathrm{}\frac{\mathrm{e}\mathrm{}\mathrm{\tau}}{\mathrm{m}}$$
The SI unit of mobility is m^{2}Vs^{-1}. The practical unit of mobility is cm^{2}Vs^{-1}.
Mobility is always positive.
Limitations of Ohm’s law
The deviations broadly are one or more of the following types:
- V ceases to be proportional to I
- The relation between V and I depends on the sign of V. In other words, if I is the current for a certain V, then reversing the direction of V keeping its magnitude fixed, does not produce a current of the same magnitude as I in the opposite direction, example, diode.
- The relation between V and I is not unique, i.e., there is more than one value of V for the same current I. Example GaAs.
- Metals have low resistivities in the range of 10^{-8} Ω m to 10^{-6} Ω m.
- Semiconductors have ρ between 10^{-6} Ω m and 10^{5} Ω m.
- Insulator have resistivity greater than 10^{5} Ω m.
Temperature dependence of resistivity
- The resistivity of a metallic conductor is approximately given by,
ρ_{T} = ρ_{0} [1 + α (T–T_{0})]
Here, α is called the temperature co-efficient of resistivity.
- Nichrome (which is an alloy of nickel, iron and chromium), Manganin and constantan exhibit a very weak dependence of resistivity with temperature. These materials are thus widely used in wire bound standard resistors since their resistance values would change very little with temperatures.
- The resistivity of semiconductors decreases with increasing temperatures.
Explanation of temperature dependence of resistivity
Resistivity of a material is given by
$$\mathrm{\rho}=1/\mathrm{\sigma}=\frac{\mathrm{m}}{\mathrm{n}{\mathrm{e}}^{2}\mathrm{\tau}}\mathrm{}$$
- As we increase temperature, average speed of the electrons increases resulting in more frequent collisions. The average time of collisions τ, thus decreases with temperature.
- In a metal, n is not dependent on temperature to any appreciable extent and thus the decrease in the value of τ with rise in temperature causes ρ to increase.
- In alloys also, n is not dependent on temperature to any appreciable extent. The movement of electrons depends upon the free space available. The free space is considerably reduced in alloys. As a result the electrons, the movement of electrons does not increase too much and as a result, the collisions, with the core or other electrons, remain more or less same, i.e., τ does not change too much. That is why alloys have no temperature dependence of resistivity.
- For insulators and semiconductors, n increases with temperature and τ decreases. The increase in n more than compensates any decrease in τ. Thus for semiconductors and non conductors, ρ decreases with temperature.
Colour coding of carbon resistors
The resistance of a carbon resistor can be calculated by the code given on it in the form of coloured strips.
Colour |
Power of 10 |
Value |
Tolerance |
Black |
0 |
1 | |
Brown |
1 |
10^{1} | |
Red |
2 |
10^{2} | |
Orange |
3 |
10^{3} | |
Yellow |
4 |
10^{4} | |
Green |
5 |
10^{5} | |
Blue |
6 |
10^{6} | |
Violet |
7 |
10^{7} | |
Grey |
8 |
10^{8} | |
White |
9 |
10^{9} | |
Gold |
-1 |
5% | |
Silver |
-2 |
10% | |
No Colour |
20% |
Yellow = 4 , Violet = 7, Brown = 1,
Gold = 5% (tolerance)
Hence R = (47 × 10 Ω) ± 5%.
Red = 2, Silver = 10%
Hence R = (22 × 10^{2} Ω) ± 10%.
Electrical energy and power
Consider a conductor with end points A and B, in which a current I is flowing from A to B. Since current is flowing from A to B, V(A) > V(B) and the potential difference across AB is V = V(A) – V(B) > 0.
In a time interval Δt, an amount of charge ΔQ = I Δt travels from A to B. The potential energy of the charge at A, was Q V(A) and at B, it is Q V(B). Thus, change in its potential energy ΔU_{p} is
ΔU_{p} = Final potential energy – Initial potential energy
= ΔQ[(V (B) – V (A)]
= –ΔQ V = –I VΔt < 0
Change in KE = - Change in PE
⇒ ΔK = –ΔU_{p}
Since ΔU_{p }< 0, therefore, ΔK > 0 for moving electrons under the influence of electric field.
Or ΔK = I VΔt
Work done W = ΔK
= I VΔt
= I^{2}RΔt
= V^{2}Δt/R
This is called Ohmic loss.
Power P = $\frac{\mathrm{W}}{\mathrm{\Delta}\mathrm{t}}$
= IV
= I^{2}R
$$=\frac{{\mathrm{V}}^{2}}{\mathrm{R}}$$
Power dissipation in transmission lines
Consider a device of resistance R, to which a power P is to be delivered via transmission cables having a resistance R_{c.}
The connecting wires from the power station to the device have a finite resistance R_{c}. The power dissipated in the connecting wires is P_{c} = I^{2}R_{c}
Now, if V is the voltage across R and I the current through it, then
$$\mathrm{P}\mathrm{}=\mathrm{}\mathrm{V}\mathrm{}\mathrm{I}\mathrm{}$$
$$\Rightarrow \mathrm{}\mathrm{I}\mathrm{}=\frac{\mathrm{P}}{\mathrm{V}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$
$$\Rightarrow \mathrm{}{\mathrm{P}}_{\mathrm{c}}\mathrm{}=\frac{{\mathrm{P}}^{2}{\mathrm{R}}_{\mathrm{c}}}{{\mathrm{V}}^{2}}\mathrm{}$$
Or power dissipated is inversely proportional to V^{2}. This is the reason the power is transmitted at high voltage.
COMBINATION OF RESISTORS
Equivalent resistance
A single resistance which can replace the combination of resistances in such a manner that the current in the circuit remains unchanged, is called the equivalent resistance.
Resistances in series
- Equivalent resistance, R = R_{1} + R_{2} + R_{3}
- Current through each resistor is same. I = I_{1} = I_{2} = I_{3}
- Sum of potential differences across individual resistors is equal to the potential difference, applied by the source. V = V_{1} + V_{2} + V_{3}
Resistances in parallel
- Equivalent resistance
$$\frac{1}{\mathrm{R}}=\frac{1}{{\mathrm{R}}_{1}}+\frac{1}{{\mathrm{R}}_{2}}+\frac{1}{{\mathrm{R}}_{3}}$$
- Potential difference across each resistor is same.
V = V_{1} = V_{2} = V_{3}
- Sum of electric currents flowing through individual resistors is equal to the electric current drawn from the source.
$$\mathrm{I}\mathrm{}=\mathrm{}{\mathrm{I}}_{1}+{\mathrm{I}}_{2}\mathrm{}+\mathrm{}{\mathrm{I}}_{3}$$
Electro motive force (emf) of a cell
Emf ε is the potential difference between the positive and negative electrodes in an open circuit, i.e., when no current is flowing through the cell.
Its SI unit is volt.
Internal resistance of a cell
The obstruction offered by the electrolyte of a cell in the path of electric current is called internal resistance (r) of the cell.
Internal resistance of a cell
- Increases with increase in concentration of the electrolyte.
- Increases with increase in distance between the electrodes.
- Decreases with increase in area of electrodes dipped in electrolyte.
Relation between ε, V and r
ε = V + Ir
ε = IR + Ir
$$\mathrm{I}\mathrm{}=\frac{\mathrm{\epsilon}}{\mathrm{R}+\mathrm{r}}$$
$$\mathrm{r}\mathrm{}=\mathrm{}\left(\frac{\mathrm{\epsilon}}{\mathrm{V}}\u2013\mathrm{}1\right)\mathrm{R}$$
If cell is in charging state, then
ε = V – Ir
Grouping of cells in series
If n cells, each of emf ε_{1}, ε_{2}, ε_{3} etc and internal resistance r_{1}, r_{2}, r_{3} etc. are connected in series to a resistance R. then equivalent emf
ε_{eq} = ε_{1} + ε_{2} + .... + ε_{n}
If ε_{1} = ε_{2} = ε_{3}…. = ε_{n} then
ε_{eq} = nε
Equivalent internal resistance
r_{eq} = r_{1} + r_{2} + r_{3} +….+ r_{n}
If r_{1} = r_{2} = r_{3}…. = r_{n}, then
r_{eq} = nr
Current in the circuit,
$$\mathrm{I}=\frac{{\mathrm{\epsilon}}_{\mathrm{e}\mathrm{q}}}{\mathrm{R}\mathrm{}+\mathrm{}{\mathrm{r}}_{\mathrm{e}\mathrm{q}}}=\frac{\mathrm{n}}{\mathrm{R}\mathrm{}+\mathrm{}\mathrm{n}\mathrm{r}}$$
Grouping of cells in parallel
If V is the potential difference between A and C, we have,
$${\mathrm{I}}_{1}\mathrm{}=\frac{({\mathrm{\epsilon}}_{1\mathrm{}}\u2013\mathrm{}\mathrm{V})}{{\mathrm{r}}_{1}}$$
$${\mathrm{I}}_{2}\mathrm{}=\frac{({\mathrm{\epsilon}}_{2\mathrm{}}\u2013\mathrm{V})}{{\mathrm{r}}_{2}}$$
Also
$$\mathrm{I}\mathrm{}=\mathrm{}{\mathrm{I}}_{1}\mathrm{}+\mathrm{}{\mathrm{I}}_{2}$$
$$=\left(\frac{{\mathrm{\epsilon}}_{1\mathrm{}}}{{\mathrm{r}}_{1}}+\frac{{\mathrm{\epsilon}}_{2\mathrm{}}}{{\mathrm{r}}_{2}}\right)-\mathrm{V}\left(\frac{1}{{\mathrm{r}}_{1}}+\frac{1}{{\mathrm{r}}_{2}}\right)$$
$$=\mathrm{}\frac{{\mathrm{\epsilon}}_{1}\mathrm{}{\mathrm{r}}_{2}+{\mathrm{\epsilon}}_{2}\mathrm{}{\mathrm{r}}_{1}}{{\mathrm{r}}_{1}{\mathrm{r}}_{2}}-\mathrm{V}\left(\frac{{\mathrm{r}}_{1}+{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}{\mathrm{r}}_{2}}\right)$$
$$\Rightarrow \mathrm{I}\mathrm{}{\mathrm{r}}_{1}{\mathrm{r}}_{2}={\mathrm{\epsilon}}_{1}\mathrm{}{\mathrm{r}}_{2}+{\mathrm{\epsilon}}_{2}\mathrm{}{\mathrm{r}}_{1}-\mathrm{}\mathrm{V}({\mathrm{r}}_{1}+{\mathrm{r}}_{2})\mathrm{}$$
Or
$$\mathrm{V}=\frac{{\mathrm{\epsilon}}_{1}\mathrm{}{\mathrm{r}}_{2}+{\mathrm{\epsilon}}_{2}\mathrm{}{\mathrm{r}}_{1}}{{\mathrm{r}}_{1}+{\mathrm{r}}_{2}}-\mathrm{}\mathrm{I}\left(\frac{{\mathrm{r}}_{1}{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}+{\mathrm{r}}_{2}}\right)$$
But
$$\mathrm{V}\mathrm{}=\mathrm{}{\mathrm{\epsilon}}_{\mathrm{e}\mathrm{q}\mathrm{}}\mathrm{}-\mathrm{I}\mathrm{}{\mathrm{r}}_{\mathrm{e}\mathrm{q}}$$
Comparing the two equations, we get,
$${\mathrm{\epsilon}}_{\mathrm{e}\mathrm{q}\mathrm{}}=\frac{{\mathrm{\epsilon}}_{1}\mathrm{}{\mathrm{r}}_{2}+{\mathrm{\epsilon}}_{2}\mathrm{}{\mathrm{r}}_{1}}{{\mathrm{r}}_{1}+{\mathrm{r}}_{2}}$$
And
$${\mathrm{r}}_{\mathrm{e}\mathrm{q}}=\mathrm{}\frac{{\mathrm{r}}_{1}{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}+{\mathrm{r}}_{2}}=\frac{1}{{\mathrm{r}}_{1}}+\frac{1}{{\mathrm{r}}_{2}}$$
$$\frac{{\mathrm{\epsilon}}_{\mathrm{e}\mathrm{q}\mathrm{}}}{{\mathrm{r}}_{\mathrm{e}\mathrm{q}}}=\frac{{\mathrm{\epsilon}}_{1\mathrm{}}}{{\mathrm{r}}_{1}}+\frac{{\mathrm{\epsilon}}_{2\mathrm{}}}{{\mathrm{r}}_{2}}\mathrm{}$$
For n cells,
$$\frac{{\mathrm{\epsilon}}_{\mathrm{e}\mathrm{q}\mathrm{}}}{{\mathrm{r}}_{\mathrm{e}\mathrm{q}}}=\frac{{\mathrm{\epsilon}}_{1\mathrm{}}}{{\mathrm{r}}_{1}}+\frac{{\mathrm{\epsilon}}_{2\mathrm{}}}{{\mathrm{r}}_{2}}+\frac{{\mathrm{\epsilon}}_{3}}{{\mathrm{r}}_{3}}+\mathrm{}\dots \frac{{\mathrm{\epsilon}}_{\mathrm{n}\mathrm{}}}{{\mathrm{r}}_{\mathrm{n}}}$$
$$\frac{1}{{\mathrm{r}}_{\mathrm{e}\mathrm{q}}}=\frac{1}{{\mathrm{r}}_{1}}+\frac{1}{{\mathrm{r}}_{2}}+\frac{1}{{\mathrm{r}}_{3}}+\mathrm{}\dots \frac{1}{{\mathrm{r}}_{\mathrm{n}}}$$
$$\Rightarrow \mathrm{}\mathrm{}{\mathrm{r}}_{\mathrm{e}\mathrm{q}}=\frac{\mathrm{r}}{\mathrm{n}}$$
If the negative terminal of the second cell is connected to positive terminal of the first cell, the equations still hold would still be valid, by replacing
$${\mathrm{\epsilon}}_{2}\to -{\mathrm{\epsilon}}_{2}$$
Kirchhoff’s laws
There are two Kirchhoff’s laws for solving complicated electrical circuits.
Junction rule
The algebraic sum of all currents meeting at a junction in a closed circuit is zero, i.e.,
Σ I = 0
This law follows law of conservation of charge.
Loop rule
The algebraic sum of all the potential differences in any closed circuit is zero, i.e.,
Σ V = 0
⇒ Σ ε = Σ IR
This law follows law of conservation of energy.
Point to be noted for loop rule
- Drop in the direction of current across each resistance is taken as negative. Drop in the direction opposite to that of current across each resistance is taken as positive.
- Value of voltage source is taken as positive if the current enter through the negative terminal (or exists from the positive terminal). Value of voltage source is taken as negative if the current enter through the positive terminal (or exits through the negative terminal)
For example, in the circuit in given above,
At junction ‘a’ the current leaving is I_{1} + I_{2} and current entering is I_{3}. Therefore as per the junction rule,
I_{3} = I_{1} + I_{2}
For the loop ‘ahdcba’ the loop rule gives,
–30 I_{1} – 41 I_{3} + 45 = 0
For the loop ‘ahdefga’ the loop rule gives,
–30 I_{1} + 21 I_{2} – 80 = 0
Balanced Wheatstone bridge
Wheatstone bridge is an arrangement of four resistances in which one resistance is unknown and the rest are known.
The bridge is said to be balanced when deflection in galvanometer is zero, i.e., I_{g} = 0.
Principle of Wheatstone bridge
$$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}}$$
If we know the value of P, Q and R, the value of unknown resistance S can be found. The bridge is most sensitive, when all the four resistances are of the same order.
Meter bridge
This is the simplest form of Wheatstone bridge and is especially useful for comparing resistance more accurately.
$$\frac{\mathrm{R}}{\mathrm{S}}=\frac{{\mathrm{l}}_{1}}{100\mathrm{}\u2013{\mathrm{l}}_{1}}$$
where l_{1} is the length of wire from one end where null point is obtained.
Potentiometer
Potentiometer is an ideal device to measure the potential difference between two points. It consists of a long resistance wire AB of uniform cross section in which a steady direct current is set up by means of a standard battery ε_{o}.
The current I through the wire can be varied by a variable resistance (rheostat, R_{h}) in the circuit.
Potential gradient
Potential drop per unit length along the potentiometer wire is called the potential gradient. It is represented by ϕ.
Taking standard symbols,
ε_{o} = emf of battery
r = internal resistance of the battery
R_{o} = resistance inserted by means of rheostat R_{h}
R = total resistance of the potentiometer wire
and I = current through the potentiometer wire
Then the potential across the wire will be,
V = IR
But I = $\frac{{\mathrm{\epsilon}}_{\mathrm{o}}}{\left({\mathrm{R}}_{\mathrm{o}}\mathrm{}+\mathrm{}\mathrm{R}\mathrm{}+\mathrm{}\mathrm{r}\right)}$
Therefore,
$$\mathrm{V}\mathrm{}=\frac{{\mathrm{\epsilon}}_{\mathrm{o}}}{\left({\mathrm{R}}_{\mathrm{o}}\mathrm{}+\mathrm{}\mathrm{R}\mathrm{}+\mathrm{}\mathrm{r}\right)}\times \mathrm{R}$$
And
$$\mathrm{}=\frac{\mathrm{V}}{\mathrm{L}}=\frac{{\mathrm{\epsilon}}_{\mathrm{o}}}{\left({\mathrm{R}}_{\mathrm{o}}\mathrm{}+\mathrm{}\mathrm{R}+\mathrm{r}\right)}\frac{\mathrm{R}}{\mathrm{L}}$$
Determination of emf of a cell using potentiometer
If with a cell of emf ε_{1}, on sliding the contact point we obtain zero deflection in galvanometer G, when contact point is at J at a length l_{1} from the end where positive terminal of cell have been joined, then fall in potential along length l_{1}, is just balancing the emf of cell. Thus, we have,
$${\mathrm{\epsilon}}_{1}\mathrm{}=\mathrm{}\mathrm{}{\mathrm{l}\mathrm{}}_{1}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$
Similarly if the balance point comes at l_{2}, when a cell of emf ε_{2} is connected, we have,
$${\mathrm{\epsilon}}_{2}\mathrm{}=\mathrm{}\mathrm{}{\mathrm{l}\mathrm{}}_{2}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$
Therefore, we can write,
$$\mathrm{}\Rightarrow \mathrm{}\frac{{\mathrm{\epsilon}}_{1}}{{\mathrm{\epsilon}}_{2}}=\frac{{\mathrm{l}}_{1}}{{\mathrm{l}}_{2}}$$
If one of the cells is a standard cell, we can find the emf of other cell.
Determination of internal resistance of a cell using potentiometer
For this, the cell (emf ε) whose internal resistance (r) is to be determined is connected across a resistance box through a key k_{2}, as shown in the figure.
If the cell E is in open circuit (k_{2} open) and balancing length l_{1}, then
ε = ϕ l_{1 }
When key k_{2} is closed the balance is obtained at l_{2},
Then
V = ϕ l_{2, }
Combining both the equations, we get,
$$\frac{\mathrm{\epsilon}}{\mathrm{V}}=\frac{{\mathrm{l}}_{1}}{{\mathrm{l}}_{2}}$$
But, ε = I (r + R) and V = IR.
$$\Rightarrow \mathrm{}\frac{\mathrm{\epsilon}}{\mathrm{V}}=\frac{\mathrm{r}+\mathrm{R}}{\mathrm{R}}$$
$$\Rightarrow \mathrm{}\frac{\mathrm{r}+\mathrm{R}}{\mathrm{R}}=\mathrm{}\frac{{\mathrm{l}}_{1}}{{\mathrm{l}}_{2}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$
$$\Rightarrow \mathrm{}\mathrm{r}\mathrm{}=\mathrm{}\mathrm{R}\left(\frac{{\mathrm{l}}_{1}}{{\mathrm{l}}_{2}}\mathrm{}-1\right)\mathrm{}$$
The potentiometer has the advantage that it draws no current from the voltage source being measured and it is unaffected by the internal resistance of the source.