NMRC JE ELECTRICAL SOLVED PAPER 2017

A series circuit is the simplest type of electrical circuit In a series circuit, there is only one path for current to flow. An electric current is a movement of charges between two points, produced by the applied voltage.

As shown in the figure the battery supplies the potential difference that forces free electrons to drift from the negative terminal at A, toward B through the connecting wires and resistances R₃, R₂, and R₁, back to the positive battery terminal at J. At the negative battery terminal, its negative charge repels electrons. Therefore, free electrons in the atoms of the wire at this terminal are repelled from A toward B. Similarly, free electrons at point B can then repel adjacent electrons, producing an electron drift toward C and away from the negative battery terminal.

At the same time, the positive charge of the positive battery terminal attracts free electrons, causing electrons to drift toward I and J. As a result, the free electrons inR₁ R₂, and R₃ are forced to drift toward the positive terminal.

The positive terminal of the battery attracts electrons just as much as the negative side of the battery repels electrons. Therefore, the motion of free electrons in the circuit starts at the same time and the same speed in all parts of the circuit.

The free electrons moving away from one point are continuously replaced by free electrons flowing to an adjacent point in the series circuit. All electrons have the same speed as those leaving the battery. An equal number of electrons move at one time with the same speed. That is why the current is the same in all parts of the series circuit.

The speed of DC motor is directly proportional to the back EMF and inversely proportional to the flux.

$N \propto \dfrac{{{E_b}}}{\Phi} = \dfrac{{V – {I_a}{R_a}}}{\Phi}$

E_b depends upon supply voltage and armature voltage drop. In case of motors without any external resistance in the armature circuit, the armature voltage drops are negligible. So, in their case, E_b is mainly dependent upon V and many times when the armature voltage drop is not known, E_b is taken equal to V.

A circuit is said to be in resonance when the reactive component or wattless component Isinφ is equal to zero. Consider a parallel circuit consisting of an inductance coil and a capacitor as shown in Fig.

Let the current through the inductance L be I_L and that through the capacitance C be Ic. if the values of L and C are such that I_L = I_C  then the resultant current I will be equal to I_R, i.e. the current flowing through the resistance R. If the circuit consists of only L and C, the currents will be only I_L and I_C, which oppose each other. Since the induction coil consists of a small but finite resistance R, the current through the inductive circuit will be

${I_R} = \dfrac{V}{Z} = \dfrac{V}{{\sqrt {\left( {{R^2} + {X^2}_L} \right)} }}$

Normally, φ is nearly equal to 90° since the value of R is very small compared to X_L. Also, the current through C is I_C = V/X_C = V/2πfC which leads V by 90°. The above circuit can be said to be in resonance if the resultant circuit current I is in phase with V. Then from the phasor diagram

I_C = I_L sinφ

${I_L} = \dfrac{V}{Z};\sin \Phi = \dfrac{{{X_L}}}{Z};{I_c} = \dfrac{V}{{{X_c}}}$

 

Putting the value

$\begin{array}{l}{I_L} = \dfrac{V}{Z};\sin \Phi = \dfrac{{{X_L}}}{Z};{I_c} = \dfrac{V}{{{X_c}}}\\\\\dfrac{V}{{{X_c}}} = \dfrac{V}{Z} \times \dfrac{{{X_L}}}{Z}\\\\{X_L}{X_c} = {Z^2}\\\\{X_c} = \dfrac{1}{{\omega C}};{X_L} = \omega L\\\\\dfrac{{\omega L}}{{\omega C}} = {Z^2}\\\\Z = \sqrt {\dfrac{L}{C}} \end{array}$

 

Since the wattless component is zero, therefore, the line current will be

line current I_r = I_L cosϕ or

 

For Resistance P the equation is given as

 

$P = \dfrac{{AB}}{{A + B + C}}$

For resistance Q the equation is

$Q = \dfrac{{AC}}{{A + B + C}}$

For resistance R

$R = \dfrac{{BC}}{{A + B + C}}$

Hence

$P = \dfrac{{20 \times 40}}{{20 + 40 + 40}} = 8\Omega$

$Q = \dfrac{{40 \times 40}}{{20 + 40 + 40}} = 16\Omega$

$R = \dfrac{{20 \times 40}}{{20 + 40 + 40}} = 8\Omega$

A 2000/200 V transformer is one that has normal primary and secondary voltages of 2000 V and 200V respectively.

Hence transformer ratio K is given as

$\begin{array}{l}K = \dfrac{{{E_2}}}{{{E_1}}} = \dfrac{{{N_2}}}{{{N_1}}}\\\\ = \dfrac{{200}}{{2000}} = \dfrac{{66}}{{{N_1}}}\\\\{N_1} = 660\end{array}$

The energy meter is an integrating meter which measures the electrical energy consumed by a load. Induction type energy meters are very commonly used to measure the electrical energy consumed in domestic, commercial and industrial installations. The commercial unit of energy is kilowatt-hour. Hence these meters measure energy in kWh.

• Open circuit test or no load test on a transformer is performed to determine ‘no-load loss (core loss)’ and ‘no-load current I₀‘ in the transformer.

• At load, the primary current has two components: Shunt branch current & secondary current when referred to the primary side.

• At no load, the secondary current becomes zero due to open circuited at secondary.

• Therefore, only shunt branch component is left which is used to magnetize the transformer coils. The current flowing in coils leads to winding losses. Hence, the current supplied by the source at no load is shunt branch current which is known as No Load Current.

• During OC test we will give rated supply(220 V here) to Low Voltage side keeping the High voltage side open. It is because here the current flowing will be no load current which is only a fraction of full load current.
• For this reason, we can neglect the variable copper losses and whatever power is injected that may be considered as core losses.(I recommend you to refer to the expression of core losses i. e hysteresis (Ph) and eddy current losses (Pi) which depends upon applied voltage & frequency and applied voltage respectively). So in OC test, we are getting the approximate rated voltage core losses.
• No load current has two components (a)Iron loss component (b) Magnetizing Component.
• Iron loss component: It is the component of no-load current responsible for the resistive loss in the core.
• Magnetizing Component: It is the component of no-load current responsible for hysteresis loss in the core
• In this method, the secondary of the transformer is left open-circuited.
• A wattmeter is connected to the primary.
• An ammeter is connected in series with the primary winding. A voltmeter is optional since the applied voltage is the same as the voltmeter reading.
• Rated voltage is applied at primary.
• If the applied voltage is normal voltage then normal flux will be set up. Since iron loss is a function of applied voltage, the normal iron loss will occur. Hence the iron loss is maximum at rated voltage. This maximum iron loss is measured using the wattmeter.

 

In the isolation transformer, the number of winding in primary and secondary is the same i.e ratio of the winding is in 1:1

In the step-up transformer, the secondary winding is more than the primary

In the step-down transformer, the primary winding is more than the secondary winding.

Hence option 1 is correct for an Isolation transformer

Option B is for step up transformer

Option C is for step down transformer

Option D is for autotransformer

A Universal motor can also refer as the AC series motor which can work on both AC and DC supply. Its construction is very similar to DC series motor.

In a normal dc. motor if the direction of both field and armature current is reversed, the direction of torque remains unchanged. So when the normal d.c.series motor is connected to an a.c supply, both field and armature current get reversed and unidirectional torque gets produced in the motor hence motor can work on a.c. supply.

But the performance of such motor is not satisfactory due to the following reasons:

1. There are tremendous eddy current losses in the yoke and field cores, which causes overheating.
2. Armature and field winding offer high reactance to a.c. due to which operating power factor is very low.
3. The sparking at brushes is a major problem because of high voltage and current.

Now to compensate for increased in the armature reaction, it is necessary to use compensating winding. The flux produced by this winding it opposite to that produced by the armature and effectively neutralizes the armature reaction.

Conductive Compensated Winding

If such a compensating winding is connected in series with the armature the motor is said to be ‘conductively compensated‘. For motors to be operated on a.c. and d.c. both, the compensation should be conductive.

In the conductively compensated type of motor, an additional compensating winding is placed in slots cut directly into the pole faces. The strength of this field increases with an increase in load current and thus minimizes the distortion of the main field flux by the armature flux called armature reaction,. The compensating winding is connected in series with the series field winding and the armature. Although conductively compensated motors have a high starting torque, the speed regulation is poor. A wide range of speed control is possible with the use of resistor-type starter-controllers.

Inductive Compensated Winding

If compensating winding is short-circuited on itself the motor is said to be ‘inductively compensated‘. In this compensating winding acts as a secondary of transformer and armature as its primary.

This winding is placed so that it links the cross-magnetizing flux of the armature, which acts as the primary winding of a transformer. Because the magnetomotive force of the secondary is nearly opposite in phase and equal in magnitude to the primary magnetomotive force, the compensating winding flux nearly neutralizes the armature cross flux. This type of motor cannot be used in DC. Because of its dependency on induction, the operating characteristics of an inductively compensated motor art very similar to those of the conductively compensated motor.

3 point starters in DC Shunt and Compound machines serve for following purposes.

• It limits the high starting current into the armature by having the resistance high at the time of starting and reducing it during running condition.
• It also protects the motor from overload and under voltage conditions.