EMF Equation Concept of a DC Generator

EMF EQUATION CONCEPT OF DC GENERATOR

Emf and Internal Resistance

Now, real batteries are constructed from materials which possess non-zero resistivities. It follows that real batteries are not just pure voltage sources. They also possess internal resistances. Incidentally, a pure voltage source is usually referred to as an emf (which stands for electromotive force). Of course, emf is measured in units of volts. A battery can be modeled as an emf connected in series with a resistor , which represents its internal resistance. Suppose that such a battery is used to drive a current through an external load resistor , as shown in Fig. 17. Note that in circuit diagrams an emf is represented as two closely spaced parallel lines of unequal length. The electric potential of the longer line is greater than that of the shorter one by volts. A resistor is represented as a zig-zag line.

Figure 17: A battery of emf and internal resistance connected to a load resistor of resistance .

Consider the battery in the figure. The voltage of the battery is defined as the difference in electric potential between its positive and negative terminals: i.e., the points and , respectively. As we move from to , the electric potential increases by volts as we cross the emf, but then decreases by volts as we cross the internal resistor. The voltage drop across the resistor follows from Ohm's law, which implies that the drop in voltage across a resistor , carrying a current , is in the direction in which the current flows. Thus, the voltage of the battery is related to its emf and internal resistance via

(133)

Now, we usually think of the emf of a battery as being essentially constant (since it only depends on the chemical reaction going on inside the battery, which converts chemical energy into electrical energy), so we must conclude that the voltage of a battery actually decreases as the current drawn from it increases. In fact, the voltage only equals the emf when the current is negligibly small. The current draw from the battery cannot normally exceed the critical value

(134)

since for I_0$" align="MIDDLE" border="0" width="55" height="35"> the voltage becomes negative (which can only happen if the load resistor is also negative: this is essentially impossible). It follows that if we short-circuit a battery, by connecting its positive and negative terminals together using a conducting wire of negligible resistance, the current drawn from the battery is limited by its internal resistance. In fact, in this case, the current is equal to the maximum possible current .

A real battery is usually characterized in terms of its emf (i.e., its voltage at zero current), and the maximum current which it can supply. For instance, a standard dry cell (i.e., the sort of battery used to power calculators and torches) is usually rated at and (say) . Thus, nothing really catastrophic is going to happen if we short-circuit a dry cell. We will run the battery down in a comparatively short space of time, but no dangerously large current is going to flow. On the other hand, a car battery is usually rated at and something like (this is the sort of current needed to operate a starter motor). It is clear that a car battery must have a much lower internal resistance than a dry cell. It follows that if we were foolish enough to short-circuit a car battery the result would be fairly catastrophic (imagine all of the energy needed to turn over the engine of a car going into a thin wire connecting the battery terminals together).

EMF EQUATION & ARMATURE RESISTANCE

We shall now derive an expression for the e.m.f. generated in a d.c. generator.

Let

Ø= flux/pole in Wb
Z = total number of armature conductors
P = number of poles
A = number of parallel paths = 2 … for wave winding
= P … for lap winding
N = speed of armature in r.p.m.
Eg= e.m.f. of the generator = e.m.f./parallel path

Armature Resistance (Ra)

The resistance offered by the armature circuit is known as armature resistance (Ra) and includes:
(i) resistance of armature winding
(ii) resistance of brushes
The armature resistance depends upon the construction of machine. Except for small machines, its value is generally less than 1Ω.

VOLTAGE & POWER EQUATION OF DC MOTOR

DC MOTOR

V = applied voltage
E_b = back e.m.f.
R_a = armature resistance
I_a = armature current

Since back e.m.f. Eb acts in opposition to the applied voltage V, the net voltage across the armature circuit is V- E_b. The
armature current I_a is given by;

I_a = (V – E_b)/ R_a

or V = E_{b +} I_aR_{a ……………………………..}(i)

This is known as voltage equation of the d.c. motor.

Power Equation

If Eq.(i) above is multiplied by I_a throughout, we get,

VI_a = E_bI_{a +}I²_aR_a

VI_a= electric power supplied to armature (armature input)
E_bI_a = power developed by armature (armature output)
I²_aR_a = electric power wasted in armature (armature Cu loss)

Thus out of the armature input, a small portion (about 5%) is wasted as a I²_aRa and the remaining portion E_bI_a is converted into mechanical power within the armature.

Condition For Maximum Power

The mechanical power developed by the motor is P_m= E_bI_a

Now P_m=VI_{a -}I²_aR_a

Since, V and R_a are fixed, power developed by the motor depends upon armature current. For maximum power, dP_m/dI_a should be zero.

dP_m/dI_{a =} V – 2I_aR_a

or I_aR_{a =} V/2

Now, V = E_{b +} I_aRa =E_{b +} V/2

therefore Eb= V/2

Hence mechanical power developed by the motor is maximum when back e.m.f. is equal to half the applied voltage.

Limitations
In practice, we never aim at achieving maximum power due to the following reasons:
(i) The armature current under this condition is very large—much excess of rated current of the machine.
(ii) Half of the input power is wasted in the armature circuit. In fact, if we take into account other losses (iron and mechanical), the efficiency will be well below 50%.