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Case Study Based Questions
Case Study 1
Read the given passage carefully and give the answer of the following questions: Q1. What is the unit of capacitive reactance? a. Farad b. Ampere c. Ohm d. Ohm-^1 Q2. The capacitive reactance of a 5μF capacitor for a frequency of 10^6 Hz is: a. 0.032 Ω b. 2.52 Ω c. 1.25 Ω d. 4.51 Ω Q3. In a capacitive circuit, resistance to the flow of current is offered by: a. resistor b. capacitor c. inductor d. frequency Q4. In a capacitive circuit, by what value of phase angle does alternating current leads the emf? a. 45° b. 90° c. 75° d. 60°
Alternating Current
Q5. One microfarad capacitor is joined to a 200 V, 50 Hz alternator. The rms current through capacitor is: a. 6.28 × 10-^2 A b. 7.5 × 10-^4 A c. 10.52 x 10-^2 A d. 15.25 × 10-^2 A
Solutions
- (c) Ohm Ohm is the unit of capacitive reactance.
- (a) 0.032 Ω Capacitive reactance,
- (b) capacitor In capacitive circuit, resistance to the flow of current is offered by the capacitor.
- (b) 90°
- (a) 6.28 × 10-^2 A
Case Study 2
The large-scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-
- (d) Step-down transformer decreases the AC voltage.
- (a) the voltage is stepped-up (Vs > Vp) and the arrangement is called a step-up transformer. If number of turns in secondary coil (Ns) are greater than number of turns in primary (Np), then voltage is increased or stepped-up in secondary, so the arrangement is called step-up transformer.
- (a) The current is reduced and consequently, the I^2 R loss is cut down.
- (c) 400
Case Study 3
Step-down transformers are used to decrease or step-down voltages. These are used when voltages need to be lowered for use in homes and factories. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000 - 220 V step-down transformer at a sub-station in the town.
Read the given passage carefully and give the answer of the following questions: Q1. The value of total resistance of the wire is: a. 25 Ω b. 30 Ω c. 35 Ω d. 15 Ω Q2. The line power loss in the form of heat is: a. 550 kW b. 650 kW c. 600 kW d. 700 kW Q3. How much power must the plant supply, assuming there is negligible power loss due to leakage? a. 600 kW b. 1600 kW c. 500 W d. 1400 kW Q4. The voltage drop in the power line is: a. 1700 V b. 3000 V c. 2000 V d. 2800 V Q5. The total value of voltage transmitted from the plant is: a. 500 V b. 4000 V c. 3000 V d. 7000 V
Solutions
- (d) 15 Ω Resistance of the two wire lines carrying power = 0.5 Ω /km Total resistance = (15+15) 0.5 = 15 Ω
Q3. At what time is the energy stored completely electrical? Q4. At what time is the energy stored completely magnetic? Q5. Calculate the value of XL.
Solutions
Case Study 5
When the frequency of AC supply is such that the inductive reactance and capacitive reactance become equal, the impedance of the series LCR circuit is equal to the ohmic
resistance in the circuit. Such a series LCR circuit is known as resonant series LCR circuit and the frequency of the AC supply is known as resonant frequency. Resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. We cannot have resonance in an RL or RC circuit. A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230V variable frequency supply. Read the given passage carefully and give the answer of the following questions: Q1. Find the value of source frequency for which current amplitude is maximum. Q2. Calculate the value of maximum current. Q3. Calculate the value of maximum power. Q4. At resonance, which physical quantity is maximum?
Solutions
**6.
9.**
**10.
11.**
