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State two; advantages of parallel connection of electric loads.(2 marks)craft1 electrical June/July 2020
If one load in a parallel circuit fails or is disconnected, the other loads will continue to function because they are connected to the power source through separate paths. The voltage across each load in a parallel circuit is the same, so the loads can operate at their optimal voltage and functionRead more
State two; a (1) disadvantages of series connection of electric loads.(2 marks)craft1 electrical June/July 2020
If one load in the series circuit fails or is disconnected, the entire circuit will be interrupted and all of the loads will stop functioning. The voltage drop across each load in a series circuit is the same, so the loads may not operate at their optimal voltage and may not function properly. The tRead more
State two practical applications of heating effect of electric current.(2 marks)craft1 electrical June/July 2020
Electric ovens: Electric ovens use heating elements, which are metal wires that are heated by the flow of electric current through them. This heat is then used to cook or bake food. Soldering irons: Soldering irons are used to join two metal pieces together by melting a small amount of metal (solderRead more
An air cored toroidal coil has 3000 turns and carries current 0.1 A. the length of the magnetic circuit is 30 cm and the cross-sectional area of the coil is 4cm^2. Determine the: (i) magnetic field strength; (ii) magnetic flux density. (6 marks)craft1 electrical June/July 2020
i. To find the magnetic field strength, we can use the formula B = u0 * n * I / l, where B is the magnetic field strength, u0 is the permeability of free space, n is the number of turns in the coil, I is the current in the coil, and l is the length of the magnetic circuit. Plugging in the values givRead more
i. To find the magnetic field strength, we can use the formula B = u0 * n * I / l, where B is the magnetic field strength, u0 is the permeability of free space, n is the number of turns in the coil, I is the current in the coil, and l is the length of the magnetic circuit. Plugging in the values given, we get B = (4pi10^-7 Tm/A) * 3000 turns * 0.1 A / 30 cm = 4.210^-4 T.
ii. To find the magnetic flux density, we can use the formula B = u0 * n * I / (A * l), where B is the magnetic flux density, u0 is the permeability of free space, n is the number of turns in the coil, I is the current in the coil, A is the cross-sectional area of the coil, and l is the length of the magnetic circuit. Plugging in the values given, we get B = (4pi10^-7 Tm/A) * 3000 turns * 0.1 A / (4 cm^2 * 30 cm) = 3.110^-3 T.
See lessDefine the following terms as used in magnetism: (i) reluctance; (ii) hysteresis. (5 marks)craft1 electrical June/July 2020
(i) Reluctance is a measure of the resistance of a material to being magnetized. It is defined as the ratio of the magnetomotive force (MMF) required to produce a certain level of magnetization in the material to the resulting magnetization. The unit of reluctance is the ampere-turn per weber (At/WbRead more
(i) Reluctance is a measure of the resistance of a material to being magnetized. It is defined as the ratio of the magnetomotive force (MMF) required to produce a certain level of magnetization in the material to the resulting magnetization. The unit of reluctance is the ampere-turn per weber (At/Wb).
(ii) Hysteresis is the lag in the response of a material to changes in an applied magnetic field. It is typically observed as a loop on a graph of magnetization versus applied magnetic field, and it is a measure of the energy loss associated with reversing the magnetization of the material. In a material with hysteresis, it takes more energy to magnetize the material in one direction than it does to demagnetize it, and vice versa. This phenomenon is important in the design of devices such as transformers and motors, as it can result in losses due to the hysteresis loop.
See lessA technician at a battery manufacturing plant has a liquid bottle labelled “inorganic. acid”. Outline four chemical methods used to verify the properties of the liquid. (4 marks)craft1 electrical June/July 2020
Density measurement: The density of a liquid can be measured using a hydrometer or a digital density meter. This can help to confirm the identity of the liquid, as different inorganic acids have different densities. pH measurement: The pH of a liquid can be measured using a pH meter or pH paper. InoRead more
State Fleming’s right hand rule for electric generators. (3 marks)craft1 electrical June/July 2020
Fleming's right hand rule for electric generators states that if the thumb, forefinger, and middle finger of the right hand are extended and mutually perpendicular, with the forefinger pointing in the direction of the magnetic field and the thumb pointing in the direction of motion of the conductor,Read more
Fleming’s right hand rule for electric generators states that if the thumb, forefinger, and middle finger of the right hand are extended and mutually perpendicular, with the forefinger pointing in the direction of the magnetic field and the thumb pointing in the direction of motion of the conductor, the middle finger will point in the direction of the induced current.
See less(i) Define each of the following as used in simple machine: i. mechanical advantage; ii. velocity ration; iii. efficiency.(3 marks)craft1 electrical June/July 2020
(i) Mechanical advantage: Mechanical advantage is the ratio of the force applied to a simple machine to the force output by the machine. It is a measure of how much the machine amplifies the force applied to it. (ii) Velocity ratio: Velocity ratio is the ratio of the distance traveled by the input fRead more
(i) Mechanical advantage: Mechanical advantage is the ratio of the force applied to a simple machine to the force output by the machine. It is a measure of how much the machine amplifies the force applied to it.
(ii) Velocity ratio: Velocity ratio is the ratio of the distance traveled by the input force to the distance traveled by the output force in a simple machine. It is a measure of how much the machine increases or decreases the speed of the input force.
(iii) Efficiency: Efficiency is the ratio of the work output by a simple machine to the work input to the machine. It is a measure of how much of the input work is converted into useful output work, with the remainder being lost to friction and other sources of energy loss.
See lessAn industrial electric boiler rated at 4 kW contains 612 kg of water at 80°C. To adjust boiler level to safe value, 78 kg of water at 26°C was added. Determine the: (i) new temperature of water in the boiler; (ii) time required to restore water temperature to 80°C. (7 marks)craft1 electrical June/July 2020
i. To solve this problem, we can use the formula for the specific heat capacity of water: c = Q / (m * ΔT) where c is the specific heat capacity of water, Q is the heat added to the water, m is the mass of the water, and ΔT is the change in temperature. We can rearrange this formula to solve for Q:Read more
i. To solve this problem, we can use the formula for the specific heat capacity of water:
c = Q / (m * ΔT)
where c is the specific heat capacity of water, Q is the heat added to the water, m is the mass of the water, and ΔT is the change in temperature.
We can rearrange this formula to solve for Q:
Q = c * m * ΔT
First, we need to calculate the heat added to the water when it was added to the boiler. We can do this by using the specific heat capacity of water, which is 4.186 J/g°C:
Q = 4.186 J/g°C * 78 kg * (80°C – 26°C) = 214,054 J
Next, we need to calculate the new temperature of the water in the boiler. We can do this by using the formula for the heat capacity of a substance, which is given by:
C = Q / ΔT
where C is the heat capacity of the substance, Q is the heat added to the substance, and ΔT is the change in temperature.
We can rearrange this formula to solve for ΔT:
ΔT = Q / C
The heat capacity of the water in the boiler is given by:
C = c * m = 4.186 J/g°C * (612 kg + 78 kg) = 3,003,108 J/°C
Substituting these values into the formula for ΔT, we get:
ΔT = 214,054 J / 3,003,108 J/°C = 0.07°C
The new temperature of the water in the boiler is 80°C + 0.07°C = 80.07°C
ii. Finally, we need to calculate the time required to restore the water temperature to 80°C. We can do this by using the formula:
t = Q / (P * η)
where t is the time required, Q is the heat added to the water, P is the power of the heating element, and η is the efficiency of the heating element.
Since the boiler is rated at 4 kW, the power of the heating element is 4 kW. The efficiency of the heating element is not given, so we will assume it is 100% efficient (η = 1).
Substituting these values into the formula, we get:
t = 214,054 J / (4 kW * 1) = 53,513 s
The time required to restore the water temperature to 80°C is approximately 53,513 seconds, or about 14.9 hours.
See lesstwo methods of reducing effect of noise to a person in a poisy workshop. (2 marks)craft1 electrical June/July 2020
Sound-absorbing materials: Installing sound-absorbing materials, such as foam panels or carpet, can help to reduce the level of noise in the workshop by absorbing sound waves. Noise-cancelling headphones: Wearing noise-cancelling headphones can help to reduce the level of noise that a person is expoRead more