**Ques.81.** After _____ number of the time constant, the transient portion reaches more than 99% of its final value.

- 3
**5✓**- 1
- 0

The time constant represents the time required for the charge to increase from zero to 63.2% of its maximum equilibrium value. This means that in a period of time equal to one time constant, the charge on the capacitor increases from zero to 0.632.

The time constant of a circuit tells about the charging and discharging rate of a circuit. The larger the time constant, the longer the charging and discharging process will take to reach the steady-state value and vice-versa.

For example, if the time constant of a circuit in 1 second (i.e. RC = 1 second), it means that time required for the capacitor voltage to rise 63.2% of its final steady-state value is 1 second. It may be noted that the capacitor is almost fully charged (99.3%) after the time equal to **5-time constants (i.e 5 RC)**.

At one TC, i.e. at one time constant, the transient term reaches 36.8 percent of its initial value.

**Ques.82.** The two-wattmeter method is used to measure the total power in a balanced circuit powered by a 415 V, 50 Hz, three-phase, balanced power supply. If both readings are 4.5 kW but have opposite signs, the total power calculated will be.

- 9.5 kW
**0 W✓**- 950 W
- 9 kW

The reading of two wattmeters can be expressed as

W_{1} = V_{L}I_{L}cos(30 + φ)

W_{2} = V_{L}I_{L}cos(30 − φ)

** When P.F reads zero (φ = 90°**)

Such a case occurs when the load consist of pure inductance or pure capacitance

W_{1} = V_{L}I_{L}cos(30 + 90°) = – V_{L}I_{L}sin30°

W_{2} = V_{L}I_{L}cos(30 − 90°) = V_{L}I_{L}sin30°

In this condition, the two wattmeter reads equal and opposite

i.e W_{1} + W_{2} = 0 = 4.5 − 45 = 0

**When PF is 0.5** (**φ = 60°)**

W_{1} = V_{L}I_{L}cos90° = 0

W_{2} = V_{L}I_{L}cos30°

Hence total power is measured by wattmeter W_{2} alone

**(iii) When PF is less than 0.5 but greater than 0 i.e ( 90° > φ > 60°)**

W_{1} = Negative

W_{2} = positive (since cos(−φ) = cosφ)

The wattmeter W_{2} reads positive (i.e.upscale) because for the given conditions (i.e. ( 90° > φ > 60°), the phase angle between voltage and current will be less than 90°.

However, in wattmeter W_{1}, the phase angle between voltage and current shall be more than 90° and hence the wattmeter gives a negative (i.e. downscale) reading.

Wattmeter cannot show a negative reading as it has only a positive scale. An indication of negative reading is that the pointer tries to deflect in a negative direction i.e. to the left of zero.

In such a case, reading can be converted to positive by interchanging either pressure coil connections or by interchanging current coil connections. Remember that interchanging connections of both the coils will have no effect on wattmeter reading.

**When PF is unity ( φ = 0°)**

W_{1} = V_{L}I_{L}cos30°

W_{2} = V_{L}I_{L}cos30°

Both wattmeters read equal and positive reading i.e upscale reading

**Ques.83.** The viscosity of the transformer oil is

- Very High
- High
**Low✓**- Medium

Transformer oil is used in the oil-filled transformer and in some other systems such as high voltage capacitors, fluorescent lamp ballasts, circuit breaker etc.

Transformers generate a lot of heat by dropping high voltage to low voltage. The heat has to be removed or the copper will melt and electrical contact will be lost.

Transformer Oil has a very good heat transfer coefficient (removes heat easily), does not conduct electricity at all (so electrical shorts won’t occur), and is selected to have a high boiling point, so it remains a liquid inside the transformer. It is also very chemically stable, so there is no breakdown over time.

**Viscosity** is the measure of the resistance of a fluid to flow. A fluid that is highly **viscous has a** high resistance (like **having** more friction) and flows slower than a **low**–**viscosity** fluid. … Honey would move slower than **water. **Thus, water is “thin”, having a low viscosity, while honey is “thick” having a high viscosity.

There are two main functions of the transformer oil:-

**Coolant**:**Insulator**: Transformer oil has great dielectric strength so it can withstand a quite high voltage, so is used as an insulator in the transformer.

The transformer oil should possess the following properties:

- High dielectric strength.
- Low viscosity to provide good heat transfer.
**Low Volatility**for low vaporization of oil.- Good resistance to the emulsion.
- Free from inorganic acid, alkali, and corrosive sulfur.
- High flash/fire point.

**Ques.84.** Find the odd one from the following.

**Brass✓**- Manganin
- Constantan
- Nichrome

We know that metals contain a large number of free electrons and thus electric current can pass through them easily. For example, metals offer a little resistance to the flow of free electrons through them, i.e., they have low resistance.

Substances like manganin, constantan, nichrome, tungsten, etc., offer high resistance to the flow of electric current through them. Hence, they have high resistance. Constantan (or eureka) is an alloy of copper (55%) and nickel (45%). It is used in the construction of pyrometers and thermocouples.

Manganin is an alloy of copper (83%), manganese (13%), and nickel (4%). It is used in the construction of rheostats and resistors. Nichrome is an alloy of nickel (60%) and chromium (40%). It is used in the construction of the heating element.

The brass has electrical conductivity much less than the pure metal. So brass is a conductor but not a good conductor.

**Ques.85.** A delta-connected balanced load is supplied from a three-phase balanced 400 V, 50 Hz AC supply. The readings on the two wattmeters are 970 W and 480 W respectively. Each phase of the load consists of resistance and inductance connected in series. Calculate the total active power consumed.

- 14.5 kW
- 1450 kW
**1.46 kW✓**- 145 kW

As given in the above question

W_{1} = 970 W

W_{2} = 480 W

V_{L} = 400 V

The power factor of the two wattmeters is

tanØ = √3[(W_{1} – W_{2}) / (W_{1} + W_{2})]

tanØ = √3[(970 – 480) ⁄ (970 + 480)]

Φ = 30°

**Power factor = cosφ = cos30° = 0.86**

Total power

W = W_{1} + W_{2 }= 970 + 480 = 1450W

Input power

P = √3.V_{L}.I_{L}.cosφ

1450 = √3 × 400 × I_{L }× 0.86

**I _{L} = 2.43 A**

In delta connected system line voltage is equal to the phase voltage

V_{P} = V_{L} = 400 V

The power consumed by the three-phase delta connected circuit is

**P = √3V _{L}.I_{L}.cosφ**

P = √3 × 400 × 2.43 × 0.86 = **1446.14 W **

**= 1.46 kW**

**Ques.86.** The line voltage V_{RY} in a three-phase system is equal to:-

- The algebraic sum of V
_{RN}and V_{NY} - The phasor sum of V
_{RN}and V_{YN} - The phasor difference between V
_{RN}and V_{NY} **The phasor sum of V**_{RN}and V_{NY}✓

In a three-phase system, there are two sets of voltages: (i) line voltages, and (ii) phase voltages. Similarly, there are two sets of currents: (i) line currents, and (ii) phase currents. We shall now determine the relations between these two sets of voltage and two sets of currents in both the star-connected system as well as the delta-connected system.

(1) **Star-Connected System**

Let us again assume the emf in each phase to be positive when acting from the neutral point outwards, as shown in Fig. The RMS values of the EMFs generated in the three phases are E_{RN}, E_{YN} and E_{BN}. In practice, it is the voltage between two lines or between a line conductor and the neutral point that is measured.

In a three-phase system, there are two sets of voltages we are interested in. One is the set of phase voltage and the other is the set of line voltages. In Fig. , V_{RN} is the RMS value of the voltage drop from R to That is, this is the phase voltage of phase R. Thus, V_{RN}, V_{YN}, and V_{BN} denote the set of three-phase voltage.

The term ‘line voltage’ is used to denote the voltage between two lines. Thus, V_{RY}represents line voltage between the lines R and Y. The other line voltages are V_{YB}and V_{BR}.

By applying Kirchhoff’s voltage law, we can get the magnitude and phase angle of the line voltage VRY (which is the voltage drop from R via N to Y. and can be represented by an unambiguous symbol V_{RNY}):

V_{RY} = V_{RN}+V_{NY}

This equation simply states that the voltage drop existing from R to Y is equal to the voltage drop from R to N plus the voltage drop from N to Y. The above equation can be written as

V_{RY} = V_{RN}+V_{NY} = V_{RN }– V_{YN} = V_{RN} + (−V_{YN})

Hence the line voltage V_{RY} is equal to the phasor sum of V_{RN} and V_{NY} which is also equal to the phasor difference of V_{RN} and V_{YN} (V_{NY} = −VYN). Hence, V_{RY} is found by compounding V_{RN} and V_{YN} reversed.

**Ques.87.** The phase sequence is important in determining the direction of rotation of the ____

- Dc series motor
- BLDC motor
- DC shunt motor
**AC motor✓**

**Phase Sequence:** The sequence in which the voltages in three phases reach their maximum positive values is called phase-sequence. Generally, the phase sequence is R-Y-B.

The phase sequence is important in determining the direction of rotation of a.c. motors such as induction motor, synchronous motor, Single phase ac motor it also help in used to determine the parallel operation of alternators Connecting a generator with the wrong phase sequence will result in a short circuit as the system.

**Ques.88.** In an R-C circuit, when the switch S is closed, the response_____

**Decays with the time✓**- Rises with time
- Do not vary with time
- First increases and then decreases

In an R-C circuit, when the switch** S is closed, the response decays with time.**

**Discharging of a Capacitor**

Consider the circuit consisting of a capacitor with an initial charge Q. A resistor R and a switch S. When the switch is open, there is a potential difference of Q/C across the capacitor and zero Potential difference across the resistor since I = 0.

If the switch is closed at t = 0, the capacitor begins to discharge through the resistor. At some time, during discharging, let circuit current be I and charge on capacitor q. According to Kirchhoff’s voltage law, the potential drop across the resistor (= IR) must be equal to the potential difference across the capacitor (= Q/C) i.e.,

**IR = Q/C …**

In the circuit when switch S is closed at t = 0. Since the capacitor never allows sudden changes in voltage, it will act as a short circuit at t = 0+. So, the current in the circuit at t = 0+ is V/R. As the capacitor discharges, it loses its charge at a declining rate. At the start of discharge the initial conditions of the circuit, are t = 0, i = 0 and q = Q. The voltage across the capacitor’s plates is equal to the supply voltage and Vc = Vs.

**At t = 0, the current i = V/R**

After one time constant, T has dropped by 63%of its initial value which is 1 – 0.63 = 0.37 or 37% of its initial value. As the voltage across the plates is at its highest value maximum discharge current flows around the circuit at the beginning because the discharging rate is fastest at the start and then tapers off as the capacitor loses charge at a slower rate.

Hence when switch S is closed, the response decays with the time that is the response V/R decreases with increase in time as shown in Fig. After 5 TC, the curve reaches 99 percent of its final value.

**Ques.89.** Lead in case of the lead-acid cell is known as _____

- Negative passive material
**Negative active material✓**- Positive passive material
- Positive active material

**Lead-acid Battery**

They are the oldest and most mature among the all-battery technologies. Because of their large applications, lead-acid batteries have the lowest cost of all-battery technologies.

The lead-acid battery differs from the primary cell type of battery mainly in that it may be recharged, whereas most primary cells are not normally recharged. In addition, the term storage battery is somewhat deceiving because this battery does not store electrical energy but is a source of chemical energy that produces electrical energy.

Lead in case of the **lead-acid cell is known as Negative active material.**

As the name implies, the lead-acid battery consists of a number of lead-acid cells immersed in a dilute solution of sulfuric acid. Each cell has two groups of lead plates; one set is the positive terminal and the other is the negative terminal.

The active material consists of the positive and negative plates. The grid and active material together form an electrode, which is also called a plate. In the lead-acid design, the positive plate is a lead dioxide cathode, and the negative plate is a lead anode.

**Lead Anode**

The lead anode is also known as the negative electrode in a lead-acid cell. Its active material is sponge lead, which increases the available surface area for reacting with the sulfuric acid electrolyte.

**Lead Dioxide Cathode**

The cathode is also known as the positive electrode in a lead-acid cell. The active material on the cathode is lead dioxide which is electroformed from lead oxide powder that must be pasted onto the grid.

Active materials within the battery (lead plates and sulfuric acid electrolyte) react chemically to produce a flow of direct current whenever current-consuming devices are connected to the battery terminal posts.

This current is produced by the chemical reaction between the active material of the plates (electrodes) and the electrolyte (sulfuric acid). This type of cell produces slightly more than 2 V.

Most automobile batteries contain six cells connected in series so the output voltage from the battery is slightly more than 12 V.

In addition to being rechargeable, the main advantage of the lead-acid storage battery over the dry cell battery is that the storage battery can supply current for a much longer time than the average dry cell.

**Ques.90.** The capacitor stores 0.4 C of charge at 2 V. Its capacitance is _______

- 3.2 F
- 0.8 F
- 0.4 F
**0.2 F✓**

Charged stored in the capacitor is given as

Q = CV

Where

Q = charge stored = 0.4C

C = Capacitance =?

V = voltage 2 V

0.4 = 2C

**C = 0.4/2 = 0.2 F **