Topic covered

`color{blue}{star}` ELECTRICAL ENERGY, POWER


`color{blue} ✍️` Consider a conductor with end points `A` and `B`, in which a current `I` is flowing from `A` to `B`. The electric potential at `A` and `B` are denoted by `V(A)` and `V(B)` respectively.

`color{blue} ✍️` Since current is flowing from A to B, `V(A) > V(B)` and the potential difference across AB is `color{purple}(V = V(A) – V(B) > 0.)`

`color{blue} ✍️` In a time interval `Δt`, an amount of charge `ΔQ = I Δt` travels from `A` to `B`.

`color{blue} ✍️` The potential energy of the charge at `A`, by definition, was `Q V(A)` and similarly at `B`, it is `Q V(B)`. Thus, change in its potential energy `ΔU_(pot)` is

`color{purple}(ΔU_(pot) =)` Final potential energy – Initial potential energy

`color{purple}(= ΔQ[(V (B) – V (A)] = –ΔQ V)`

`color{purple}(= –I VΔt < 0)`


`color{blue} ✍️` If charges moved without collisions through the conductor, their kinetic energy would also change so that the total energy is unchanged. Conservation of total energy would then imply that,

`color{purple}(ΔK = –ΔU_(pot))`


that is,

`color{purple}(ΔK = I VΔt > 0)`


`color{blue} ✍️` Thus, in case charges were moving freely through the conductor under the action of electric field, their kinetic energy would increase as they move.

`color{blue} ✍️` We have, however, seen earlier that on the average, charge carriers do not move with acceleration but with a steady drift velocity.

`color{blue} ✍️` This is because of the collisions with ions and atoms during transit. During collisions, the energy gained by the charges thus is shared with the atoms.

`color{blue} ✍️` The atoms vibrate more vigorously, i.e., the conductor heats up. Thus, in an actual conductor, an amount of energy dissipated as heat in the conductor during the time interval `Δt` is,

`color{purple}(ΔW = I VΔt)`


`color{blue} ✍️` The energy dissipated per unit time is the power dissipated `color{purple}(P = (ΔW)/(Δt))` and we have,

`color{purple}(P = I V)`


Using Ohm’s law `V = IR`, we get

`color{purple}(P = I^2 R = V^2/R)`


as the power loss (“ohmic loss”) in a conductor of resistance `R` carrying a current `I.`

`color{blue} ✍️` It is this power which heats up, for example, the coil of an electric bulb to incandescence, radiating out heat and light.

`color{blue} ✍️` As we have reasoned before, we need an external source to keep a steady current through the conductor. It is clearly this source which must supply this power.

`color{blue} ✍️` In the simple circuit shown with a cell (Fig.3.12), it is the chemical energy of the cell which supplies this power for as long as it can.

`color{blue} ✍️` The expressions for power, Eqs. (3.32) and (3.33), show the dependence of the power dissipated in a resistor R on the current through it and the voltage across it. Equation (3.33) has an important application to power transmission.

`color{blue} ✍️` Electrical power is transmitted from power stations to homes and factories, which may be hundreds of miles away, via transmission cables. One obviously wants to minimise the power loss in the transmission cables connecting the power stations to homes and factories. We shall see now how this can be achieved.

`color{blue} ✍️` Consider a device `R,` to which a power `P` is to be delivered via transmission cables having a resistance `R_c` to be dissipated by it finally. If `V` is the voltage across `R` and `I` the current through it, then

`color{purple}(P = VI)`


`color{blue} ✍️` The connecting wires from the power station to the device has a finite resistance `R_c.` The power dissipated in the connecting wires, which is wasted is `P_c` with

`color{purple}(P_C = I^2R_c)`



`color{blue} ✍️` from Eq. (3.32), Thus, to drive a device of power P, the power wasted in the connecting wires is inversely proportional to `V^2.`

`color{blue} ✍️` The transmission cables from power stations are hundreds of miles long and their resistance `R_c` is considerable. To reduce `P_c`, these wires carry current at enormous values of `V` and this is the reason for the high voltage danger signs on transmission lines — a common sight as we move away from populated areas.

`color{blue} ✍️` Using electricity at such voltages is not safe and hence at the other end, a device called a transformer lowers the voltage to a value suitable for use.


`color{blue} ✍️` The current through a single resistor `R` across which there is a potential difference `V` is given by Ohm’s law `I = V//R.` Resistors are sometimes joined together and there are simple rules for calculation of equivalent resistance of such combination.

`color{blue} ✍️` Two resistors are said to be in series if only one of their end points is joined (Fig. 3.13). If a third resistor is joined with the series combination of the two (Fig. 3.14), then all three are said to be in series. Clearly, we can extend this definition to series combination of any number of resistors.

`color{blue} ✍️` Two or more resistors are said to be in parallel if one end of all the resistors is joined together and similarly the other ends joined together (Fig. 3.15).

`color{blue} ✍️` Consider two resistors `R_1` and `R_2` in series. The charge which leaves `R_1` must be entering `R_2`. Since current measures the rate of flow of charge, this means that the same current I flows through `R_1` and `R_2.` By Ohm’s law:

`color{blue} ☞` Potential difference across `color{green}(R_1 = V_1 = I R_1),` and
`color{blue} ☞` Potential difference across `color{green}(R_2 = V_2 = I R_2).`

`color{blue} ✍️` The potential difference `V` across the combination is `color{purple}(V_1+V_2).` Hence,

`color{blue}(= V_1+ V_2 = I (R_1 + R_2)`


`color{blue} ✍️` This is as if the combination had an equivalent resistance Req, which by Ohm’s law is

`color{purple}(R_(eq) = V/I = (R_1 +R_2)`


`color{blue} ✍️` If we had three resistors connected in series, then similar

`color{purple}(V = I R_1 + I R_2 + I R_3 = I (R_1+ R_2+ R_3)`


`color{blue} ✍️` This obviously can be extended to a series combination of any number n of resistors `color{purple}(R_1, R_2 .....,)` Rn. The equivalent resistance `R_(eq)` is

`color{purple} (R_(eq) = R_1 + R_2 + . . . + R_n)`


`color{blue} ✍️` Consider now the parallel combination of two resistors (Fig. 3.15). The charge that flows in at `A` from the left flows out partly through `R_1` and partly through `R_2`. The currents `I, I_1, I_2` shown in the figure are the rates of flow of charge at the points indicated. Hence,

`color{purple}(I= I_1+I_2)`


`color{blue} ✍️` The potential difference between `A` and `B` is given by the Ohm’s law applied to `R_1`

`color{purple}(V = I_1 R_1)`


Also, Ohm’s law applied to `R_2` gives

`color{purple}(V = I_2 R_2)`.........(3.42)

`color{purple}( I=I_1 + I_2 = V/(R_1) + V/(R_2) = V[ (1/R_1)+1/(R_2))]`


`color{blue} ✍️` If the combination was replaced by an equivalent resistance `R_(eq,)` we would have, by Ohm’s law

`color{purple}(I = V/(R_(eq))`



`color{purple}((1/(R_(eq)) = 1/(R_1) + 1/(R_2))`


`color{blue} ✍️` We can easily see how this extends to three resistors in parallel (Fig. 3.16).

Exactly as before

`color{purple}(I = I_2+ I_2 +I_3)`


and applying Ohm’s law to `R_1, R_2` and `R_3` we get

`color{purple}(V = I_1 R_1, V = I_2 R_2, V = I_3 R_3)`


So that

`color{purple}(I = I_1 + I_2 + I_3 =V (1/R_1+1/R_2+1/R_3))`


`color{blue} ✍️` An equivalent resistance `R_(eq)` that replaces the combination, would be such that

`color{purple}(I = V/(R_(eq))`


and hence

`color{purple}((1/(R_(eq)) = 1/R_1 + 1/R_2 + 1/R_3))`


`color{blue} ✍️` We can reason similarly for any number of resistors in parallel. The equivalent resistance of n resistors `color{purple}(R_1, R_2 . . . ,R_n)` is

`color{purple}(1/R_(eq) = 1/R_1+1/R_2 + ..... + 1/R_n)`


`color{blue} ✍️` These formulae for equivalent resistances can be used to find out currents and voltages in more complicated circuits. Consider for example, the circuit in Fig. (3.17),

`color{blue} {➢➢}` where there are three resistors `color{purple}(R_1, R_2)` and `color{purple}(R_3.` `R_2)` and `R_3` are in parallel and hence we can replace them by an equivalent `R_(eq)^(23)` between point `B` and `C` with

`color{purple}(R_(eq)^(23)= 1/R_2 + 1/R_3)`


`color{purple}(R_(eq)^(23) = (R_2 R_3)/(R_2+R_3))`


`color{blue} {➢➢}` The circuit now has `R_1` and `R_(eq)^(23)` in series and hence their combination can be replaced by an equivalent resistance `R_(eq)^(23)` with

`color{purple}(R_(eq)^(123) = R_(eq)^(23) +R_1)`


`color{blue} {➢➢}` If the voltage between `A` and `C` is `V`, the current `I` is given by

`color{purple}(I = V/R_(eq)^(123) = V/(R_1+[R_2R_3//(R_2+R_3)])`




`color{blue} ✍️` We have already mentioned that a simple device to maintain a steady current in an electric circuit is the electrolytic cell.

`color{blue} ✍️` Basically a cell has two electrodes, called the positive ` (P)` and the negative `(N)`, as shown in Fig. 3.18.

`color{blue} ✍️` They are immersed in an electrolytic solution. Dipped in the solution, the electrodes exchange charges with the electrolyte.

`color{blue} ✍️` The positive electrode has a potential difference `V_+ (V_+ > 0)` between itself and the electrolyte solution immediately adjacent to it marked `A` in the figure.

`color{blue} ✍️` Similarly, the negative electrode develops a negative potential `color{purple}(– (V_(–) ) (V_(–) ≥ 0))` relative to the electrolyte adjacent to it, marked as `B` in the figure.

`color{blue} ✍️` When there is no current, the electrolyte has the same potential throughout, so that the potential difference between P and `N` is `color{purple}(V^(+) – (–V_(–)) = V_(+) + V_(–))` .

`color{blue} ✍️` This difference is called the electromotive force (emf) of the cell and is denoted by `ε`. Thus

`color{purple}(ε = V_(+)+V_(–) > 0)`


`color{brown} bbul{"Note that"}` `ε` is, actually, a potential difference and not a force. The name emf, however, is used because of historical reasons, and was given at a time when the phenomenon was not understood properly.

`color{blue} ✍️` To understand the significance of `ε,` consider a resistor `R` connected across the cell (Fig. 3.18). A current `I` flows across `R` from `C` to `D`. As explained before, a steady current is maintained because current flows from `N` to `P` through the electrolyte.

`color{blue} ✍️` Clearly, across the electrolyte the same current flows through the electrolyte but from `N` to `P`, whereas through `R`, it flows from `P` to `N`.

The electrolyte through which a current flows has a finite resistance `r`, called the internal resistance. Consider first the situation when `R` is infinite so that `I = V//R = 0,` where `V` is the potential difference between `P` and `N`. Now,

`V = `Potential difference between `P` and `A +` Potential difference between `A` and `B +` Potential difference between `B` and `N` .
`= epsilon` ...........(3.56)

`color{blue} ✍️` Thus, 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. If however R is finite, I is not zero. In that case the potential difference between `P` and `N` is

`color{purple}(V = V_(+)+ V_(–) – I r)`

`color{purple}(= ε – I r)`


`color{blue} ✍️` Note the negative sign in the expression `(I r)` for the potential difference between `A` and `B`. This is because the current `I` flows from `B` to `A` in the electrolyte.

`color{blue} ✍️` In practical calculations, internal resistances of cells in the circuit may be neglected when the current `I` is such that `ε >> I r.`

`color{blue} ✍️` The actual values of the internal resistances of cells vary from cell to cell. The internal resistance of dry cells, however, is much higher than the common electrolytic cells.

`color{blue} ✍️` We also observe that since `V` is the potential difference across `R`, we have from Ohm’s law

`color{purple}(V = I R)`

Combining Eqs. (3.57) and (3.58), we get

`color{purple}(I R = ε – I r)`


`color{purple}(I = ε/(R+r))`


`color{blue} ✍️` The maximum current that can be drawn from a cell is for `R = 0` and it is `Imax = ε//r`. However, in most cells the maximum allowed current is much lower than this to prevent permanent damage to the cell.
Q 3104091858

A network of resistors is connected to a 16 V battery with internal resistance of `1Ω,` as shown in Fig. 3.19: (a) Compute the equivalent resistance of the network. (b) Obtain the current in each resistor. (c) Obtain the voltage drops `V_(AB), V_(BC)` and `V_(CD.)`
Class 12 Chapter example 5

(a) The network is a simple series and parallel combination of resistors. First the two `4Ω` resistors in parallel are equivalent to a resistor `= [(4 × 4)/(4 + 4)] Ω = 2 Ω.`

In the same way, the `12 Ω` and `6 Ω` resistors in parallel are equivalent to a resistor of
`[(12 × 6)/(12 + 6)] Ω = 4 Ω.`
The equivalent resistance R of the network is obtained by combining these resistors `(2 Ω` and `4 Ω)` with `1 Ω` in series, that is,

`R = 2 Ω + 4 Ω + 1 Ω = 7 Ω.`

b) The total current I in the circuit is

`I = (epsilon)/(R+r) = ((16V)/(17+1)Ω) = 2A`

Consider the resistors between A and B. If `I_1` is the current in one of the `4 Ω` resistors and `I_2` the current in the other,

`I_1 × 4 = I_2 × 4`

that is, `I_1 = I_2,` which is otherwise obvious from the symmetry of
the two arms. But `I_1 + I_2 = I = 2 A.` Thus,

`I_1 = I_2 = 1 A`
that is, current in each 4 Ω resistor is 1 A. Current in 1 Ω resistor
between B and C would be 2 A.
Now, consider the resistances between C and D. If B is the current in the `12 Ω` resistor, and `I_4` in the `6 Ω` resistor,

`I_3 × 1_2 = I_4 × 6, i.e., I_4 = 2I3`

But, `I_3 + I_4 = I = 2 A`

Thus `I_3 = (2/3) A, I_4 = (4/3)A`

that is, the current in the `12 Ω` resistor is `(2//3)` A, while the current in the `6 Ω` resistor is `(4/3)` A.

(c) The voltage drop across AB is

`V_(AB) = I_1 × 4 = 1 A × 4 Ω = 4 V,`
This can also be obtained by multiplying the total current between A and B by the equivalent resistance between A and B, that is,

`V_(AB) = 2 A × 2 Ω = 4 V`
The voltage drop across BC is

`V_(BC) = 2 A × 1 Ω = 2 V`

Finally, the voltage drop across CD is

`V_(CD) = 2A xx 1 Ω = 2 V`

Finally, the voltage drop across CD is

`V_(CD) = 12 Ω × I_3 = 12 Ω × (2/3) A = 8V`

This can alternately be obtained by multiplying total current between C and D by the equivalent resistance between C and D, that is,

`V_(CD) = 2 A × 4 Ω = 8 V`

Note that the total voltage drop across AD is `4 V + 2 V + 8 V = 14 V.` Thus, the terminal voltage of the battery is `14 V,` while its emf is 16 V. The loss of the voltage `(= 2 V)` is accounted for by the internal resistance `1 Ω` of the battery `[2 A × 1 Ω = 2 V].`