Physics

Electrical Machines and Circuits

720 Questions

Electrical machines and circuits involve the principles of alternating and direct current, transformers, diodes, and power consumption. These concepts are essential for engineering and technical exams. Review these practice questions to test your knowledge of circuit analysis.

Transformers and impedanceAlternating and direct currentCircuit components and diodesPower consumption calculations

Electrical Machines and Circuits Questions

Multiple choice

The path of an electric current from the generating source back to the original source is a(n)

  1. incomplete ciruit

  2. complete circuit

  3. insulating current

  4. direct circuit
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A complete circuit is a closed loop path that allows current to flow from the source through conductors and back to the source. An incomplete circuit has a break (open circuit) preventing flow. There's no such thing as an insulating current or direct circuit.

Multiple choice

There are ___________ wires attached to an LED.

  1. two

  2. three

  3. one

  4. no wire is required

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

An LED has two leads (wires) of different lengths: the longer one is the anode (positive) and the shorter one is the cathode (negative).

Multiple choice

Ruby has connected 2 bulbs across 2 batteries in a circuit. How can she make the bulbs dimmer?

  1. Replace one of the batteries with a section of wire.

  2. Replace one of the batteries with a piece of rubber.

  3. Replace one of the bulbs with a section of wire.

  4. Disconnect the batteries.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Reducing the voltage in a circuit decreases the current flowing through the bulbs, making them dimmer. Removing a battery reduces the total voltage provided to the circuit.

Multiple choice

Why is the bulb brighter in the circuit with 2 batteries and one bulb?

  1. The flow of electricity is less.

  2. The flow of electricity is the same.

  3. The flow of electricity is more.

  4. The flow of electricity is blocked
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Brightness in a bulb is directly related to the amount of electrical current flowing through it. Adding more batteries increases the voltage, which drives more current through the bulb.

Multiple choice

In which circuit will the bulb or bulbs glow brightest?

  1. A simple circuit with one bulb and one battery.

  2. A simple circuit with 2 batteries and 2 bulbs.

  3. A simple circuit with 2 batteries and 1 bulb.

  4. A simple circuit with three bulbs and 1 battery.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Brightness is maximized when the voltage per bulb is highest. Two batteries powering a single bulb provide more voltage to that bulb than the other configurations listed.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

In the method using the transformers, assume that the ratio of the number of turns in the primary to that in secondary in the step-up transformer is $1:10$. If the power to the consumer has to be supplied at $200\ V$, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is:

  1. $200:1$
  2. $150:1$
  3. $100:1$
  4. $50:1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

A 60 volt - 10 watt bulb is operated at 100 volt - 60 Hz a.c. The inductance required is?

  1. 2.56 H

  2. 0.32 H

  3. 0.64 H

  4. 1.28 H
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

First, find the resistance of the bulb: R = V^2 / P = 60^2 / 10 = 360 ohms. The operating current is I = P / V = 10 / 60 = 1/6 A. When connected to 100V, the impedance Z = V_source / I = 100 / (1/6) = 600 ohms. Since Z^2 = R^2 + Xl^2, 600^2 = 360^2 + Xl^2. Xl^2 = 360000 - 129600 = 230400. Xl = 480 ohms. Since Xl = 2 * pi * f * L, 480 = 2 * 3.14 * 60 * L. L = 480 / 377 = 1.273 H, which rounds to 1.28 H.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

A $50\ Hz$ $AC$ current of crest value $1\ A$ flows, through the primary of transformer. If the mutual inductance between the primary and secondary be $0.5\ H$, the crest voltage induced  in the secondary is

  1. 75 V

  2. 150 V

  3. 100 V

  4. 300V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The induced crest voltage in the secondary coil is given by E_s = M * (di/dt)_max. The rate of change of current is di/dt = omega * I_0 = (2 * pi * f) * I_0. Substituting M = 0.5 H, f = 50 Hz, and I_0 = 1 A gives E_s = 0.5 * (2 * pi * 50 * 1) = 50 * pi approx 150.7 V, which rounds to 150 V.