Ques.61. A balanced three-phase star-connected load draws power from a 440V supply. The two connected wattmeters, W_{1} and W_{2}, indicates 5 kW and 1200 W respectively. Calculate the current in the circuit.

11.52 A✓

15.56 A

1.11 A

10.11 A

As given in the above question

W_{1} = 5 kW

W_{2} = 1200 W

V_{L} = 440 V

The power factor of the two wattmeters is

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

tanØ = √3[(5000 – 1200) ⁄ (5000 + 1200)]

Φ = 45°

Power factor = cosφ = cos45° = 0.707

Total power

W = W_{1} + W_{2 }= 5000 + 1200 = 6200W

Active power in star connected system

W = √3.V_{L}.I_{L}.cosφ

6200 = √3 × 440 × I_{L }× 0.707

I_{L} = 11.52 A

Ques.62. The two-wattmeter method is used to measure the input power of a three-phase induction motor. If the two wattmeter readings are 1,700 W and 1,100 W. determine the input current drawn from a 440 V, 3-phase AC supply.

0.54 A

0.32 A

4 A✓

0.525 A

As given in the above question

W_{1} = 1700 W

W_{2} = 1100 W

V_{L} = 440 V

The power factor of the two wattmeters is

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

tanØ = √3[(1700 – 1100) ⁄ (1700 + 1100)]

Φ = 20.30°

Power factor = cosφ = cos20.30° = 0.93

Total power

W = W_{1} + W_{2 }= 1700 + 1100 = 2800W

Input power

P = √3.V_{L}.I_{L}.cosφ

2800 = √3 × 440 × I_{L }× 0.93

I_{L} = 4 A

Ques.63. A three-phase, balanced, delta-connected load of (4 + j8)Ω is connected to a 400-V, 3-φ balanced power supply. Assuming that the phase sequence is RYB, determine the phase current I_{R}.

45.74∠−63.4°A

44.74∠−63.4°A✓

45.74∠63.4°A

44.74∠63.4°A

Given

Voltage = 400 V

Zph = (4 + j8)

|Zph| = (4^{2} + 8^{2})^{1/2}

|Zph| = 8.94 Ω

θ = tan^{−1}(b/a) = tan^{−1}(8/4) = 63.4⁰

= tan^{−1}(8/4)

Impedance per phase = (4+j8) Ω = 8.94∠63.4⁰Ω

Taking the line voltage V_{RY} = V∠0⁰ as a reference V_{RY} = 400∠0⁰V, V_{YB} = 400∠-120⁰V and V_{BR} = 400∠-240⁰V.

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

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

I_{P} = V_{p}/Z_{p}

I_{P} = 400∠0^{o})/(8.94∠63.4^{o} )

I_{P} = 44.74∠−63.4⁰A.

Ques.64. A standard sinusoidal current wave changes its polarity at___

Negative Value

Zero Value✓

Minimum Value

Maximum Value

Many a time, alternating voltages and currents are represented by a sinusoidal wave, or simply a sinusoid. The sinusoidal wave is generally referred to as a sine wave. Basically, an alternating voltage (current) waveform is defined as the voltage (current) that fluctuates with time periodically, with the change in polarity and direction.

Changing polarity refers to changing from positive to negative or negative to positive. For currents, it refers to the current flowing in one direction and then flowing in the opposite direction.

The period can be measured in the following different ways

1. From zero crossing of one cycle to zero crossing of the next cycle.

2. From positive peak of one cycle to the positive peak of the next cycle.

3. From the negative peak of one cycle to negative peak oil the next cycle.

Ques.65. If Q is the charge and C be the capacitance, then the energy stored in the capacitor is given by_____

Q/2C

QC/2

Q^{2}/2C✓

1/QC

The capacitance of a capacitor is defined as the measure of the charge stored by the capacitor per unit voltage; C = Q/V

Hence Q = CV

Energy stored In the charged capacitor

When a capacitor is charged by a voltage source (say battery) it stores the electric energy. If C = capacitance of the capacitor; Q = charge on capacitor and V is the potential difference across capacitor then energy stored in the capacitor is given by

U = CV^{2}/2 = Q^{2}/2 = QV/2

Ques.66. The two-wattmeter method is used to measure a three-phase power supply. If the two wattmeter readings are 2 kW and 500 W. Determine the total power of the circuit.

25 kW

250 W

2.5 kW✓

25 W

W_{1} = 2 kW = 2000 kW

W_{2} = 500 W

Total active power = W_{1} + W_{2 }= 2000 + 500 = 2.5 kW

Ques.67. According to Gauss’s law, if E is _______, the charge density in the ideal conductor is zero.

Positive

Negative

Zero✓

Unity

A conductor contains moving free charges. An isolated conductor is shown in Figure. Let an external electric field E be applied to it. A force will get applied on the charges inside the conductor. As a result, charges will get accumulated on the surface. These induced charges generate a counteracting internally induced field E_{i}. This internally induced field E_{i}opposes and cancels the externally applied field E.

Thus, there does not exist any electrostatic field inside a perfect conductor. As a result E = -∇ V = 0.

It implies that the potential is the same at every point inside the conductor. According to Gauss’s Law, the charge density inside a perfect conductor is zero if E= 0.

Therefore,

ρ_{v} = 0

The perfect conductor is characterized by the following properties:

The electric field inside and on the surface of the perfect conductor is zero E = 0.

The total charge of the conductor is distributed on its surface so that the electric field in the conductor and on its surface remains always zero.

The electrical field is always normal to the surface of the perfect conductor, and there is no electric field component tangential to its surface. Accordingly, the electric flux density D is normal to the surface and equal to the total charge on the surface of the conductor.

The electric potential inside and on the surface of the conductor is constant since E = –∇ V = 0. Thus, the surface of the conductor is equipotential.

Ques.68. What is the relationship between the line voltage and the phase voltage in a delta -connected load?

Line voltage = Phase current

Line voltage > Phase voltage

Line voltage < Phase voltage

Line voltage = Phase voltage✓

In delta connected system the line voltage is equal to the phase voltage and the Line current is √3 times of Phase Current.

Ques.69. A 10μF capacitor in series with a 1-MΩ resistor is connected across a 100-V DC supply. The initial rate of rise of voltage across the capacitor is

0.1V/s

10V/s✓

1V/s

0.01V/s

Given

Capacitor,C = 10μF = 10 × 10^{−6}

Resistance R = 1-MΩ = 1 × 10^{6}

Voltage V = 100 V

The initial rate of rise of voltage across the capacitor is given as

= 100 ⁄ (10 × 10^{−6}) × (1 × 10^{6})

= 10 V/s

Ques.70. Find the phase voltage if a three-phase star-connected system is connected to a 400V 50 Hz AC supply. Assume Z_{ph} consists of a resistance of 10Ω in series With an inductance of 0.0318 H.

400 V

230.94 V✓

230 V

110.24 V

In three phase star connected system the phase voltage is given as