When only voltage source is present then voltage across 1Ω resistor is

V = 5V

I = V/R = 5/1 = 5A

Connect a source of 2V, with internal resistance of 1Ω at AA’

Applying Nodal analysis at C and B

Solving equation 1 & 2

V_C = 1.125 Volt

V_B = 1.25 Volt

Current through resistor R

I_R = V_B/2 = 1.25/2 = 0.625 A

Ohm’s law of constant proportionality state that for a given conductor the ratio of voltage to current is constant i.e if the temperature, area, the length did not change the resulting current is directly proportional to the variation in the applied voltage. Hence the graph between voltage and current is linear.
R(constant) = V/I

★Ohms law does not apply to all material it only applied to “linear” or “Ohmic Resistor”.

For any given material at a certain given temperature, the resistance is given as

where ρ is a constant depending on the nature of the material of the conductor and is known as its specific resistance or resistivity.

Specific Resistance depends only on the temperature and material of the conductor but not on the dimensions of the conductor, on which resistance depends, and mechanical deformation such as stretching, etc. As ρ depends only on the material of a conductor at a given temperature, hence it is a characteristic constant.

Superposition theorem

In any Linear and bilateral network consisting of linear and bilateral energy sources (e.g. generators), the current flowing at any point in the network is the algebraic sum of the currents those flow with each source were considered separately, with all the remaining sources replaced at a time by their internal impedances.

The theorem is hence based on the linearity property and Ohm‘s law. The fundamental concept is that in linear impedance, for an increase in voltage across it, the current through it increases which is independent of the magnitude of the original current flowing through it.

S ∝ induced EMF

S ∝ 4.44fNΦ

∴ S ∝ f

S₁/S₂ = f₁/f₂

S₂ = S₁f₂/f₁

= 500 x 50/200 =125 KVA

The generated voltage of the alternator is

V = E – Ia(Ra + JXs)

Where E = Induced EMF

Ia = Armature current

Ra = Armature resistance

Xs = leakage reactance of the armature winding

For no-load armature current Ia = 0

∴ V = E  (δ = 0°)

So at no load, the angle between induced emf and terminal voltage is 0°

Power delivered by synchronous generator

For Pmax  sinδ = 1

δ = 90°

Pmax = VE/X

The starting torque of the synchronous motor is zero.

Consider the rotor is at rest and the stator field is rotating at the synchronous speed at 100% slip. Suppose the N pole of the rotating field in the stator is approaching at S pole of the rotor then the magnetic force will tend to turn the rotor in the direction opposite to the rotating field When the same N pole of the rotating field has just passed the rotor S pole, the magnet force will tend to pull the rotor in the same direction as the rotating field. These opposite interactions balance out, producing a zero net torque on the rotor.

The operating power factor in the case of a transformer depends upon the load supplied by the transformer. For example, if the transformer is supplying lamp load, the operating power factor is unity, whereas, in most industries, a large percentage of the load consists of an induction motor.

These machines operate with a lagging power factor hence the power factor of the transformer is lagging, Moreover, the magnitude of the lagging power factor also depends on the nature of the inductive load being supplied by the transformer Thus the transformer rating is always specified in terms of volt-ampere (VA) or Kilo-Volt-ampere KVA.