1. A workshop uses a set of power tools. The table below shows the power ratings of each tool and the hours they are used daily:
Tool |
Power Rating |
Daily Usage (hours) |
Circular Saw |
1200 W |
2 |
Drill |
800 W |
4 |
Angle Grinder |
1500 W |
1 |
If the electricity rate is $0.10 per kWh, what is the total monthly cost of operating these tools?
a) $12.00
b) $19.80
c) $21.30
d) $24.60
Answer: c) Step 1: Calculate the daily energy consumption for each tool.
Circular Saw: 1200 W x 2 hours = 2400 Wh = 2.4 kWh
Drill: 800 W x 4 hours = 3200 Wh = 3.2 kWh
Angle Grinder: 1500 W x 1 hour = 1500 Wh = 1.5 kWh
Step 2: Calculate the total daily energy consumption for all tools.
Total daily energy consumption = Energy consumed by Circular Saw + Energy consumed by Drill + Energy consumed by Angle Grinder
Total daily energy consumption = 2.4 kWh + 3.2 kWh + 1.5 kWh = 7.1 kWh
Step 3: Calculate the monthly energy consumption.
Monthly energy consumption = Total daily energy consumption * Number of days in the month
Assuming there are 30 days in the month:
Monthly energy consumption = 7.1 kWh/day x 30 days = 213 kWh
Step 4: Calculate the total monthly cost.
The electricity rate is $0.10 per kWh.
Total monthly cost = Monthly energy consumption * Electricity rate
Total monthly cost = 213 kWh x $0.10/kWh = $21.30
Therefore, the total monthly cost of operating these tools is $21.30.
2. In an electrical system, a substantial quantity of electric charge, precisely 80,000 coulombs, is transferred over a period of 30 minutes through a potential difference of 60 volts. Calculate the total amount of thermal energy produced during this process.
a) 48000 kJ
b) 4799 kJ
c) 1599 kJ
d) 2896 kJ
Answer: b) To calculate the heat produced (H) when 80,000 coulombs of charge (Q) is transferred in 30 minutes through a potential difference (V) of 60 volts, follow these steps:
1. Calculate the current (I):
I = Q/t
I = 80,000 C/30 x 60s
I = 44.4 A
2. Calculate the resistance (R) using Ohm's law:
R = V/I
R = 60 V/ 44.4 A
R = 1.35 Ω
3. Calculate the heat produced (H):
H = I^{2} x R x t
H = (44.4 A)^{2} x 1.35 Ω x (30 x 60 s)
H = 4799040.048 J
Converting to kilojoules:
H = 4799040.048 J/1000 = 4799 kJ
3. Determine the power dissipated by each resistor in the given circuit.
a) P_{1}: 48 W, P_{2}: 24 W, P_{3}: 96 W
b) P_{1}: 12 W, P_{2}: 18 W, P_{3}: 24 W
c) P_{1}: 16 W, P_{2}: 20 W, P_{3}: 36 W
d) P_{1}: 36 W, P_{2}: 24 W, P_{3}: 18 W
Answer: d) To calculate the power dissipated by each resistor in a parallel circuit connected to a 12V battery, we can use the formula for power:
P = I x V
Where:
P is the power in watts (W).
I is the current in amperes (A).
V is the voltage in volts (V).
Since all the resistors are connected in parallel to the same voltage source (12V battery), the voltage (V) is the same across all resistors. Therefore, we need to calculate the current (I) flowing through each resistor and then use that to calculate the power (P) for each resistor.
Using Ohm’s law, calculate the currents through each resistor:
I = V/R
Given:
R_{1} = 4 Ω
R_{2} = 6 Ω
R_{3} = 8 Ω
V = 12 V
For R_{1}:
I_{1} = V/R_{1}
I_{1} = 12/4 = 3 A
For R_{2}:
I_{2} = V/R_{2}
I_{2} = 12/6 = 2 A
For R_{3}:
I_{3} = V/R_{3}
I_{3} = 12/8 = 1.5 A
Now that we have the currents through each resistor, we can calculate the power dissipated by each resistor using the formula
P = I x V
So, the power dissipated by each resistor in the parallel circuit is as follows:
Power dissipated by R_{1}:
P_{1} = I_{1} x V
P_{1} = 3 x 12 = 36 W
Power dissipated by R_{2}:
P_{2} = I_{2} x V
P_{2} = 2 x 1_{2} = 24 W
Power dissipated by R_{3}:
P_{3} = I_{3} x V
P_{3} = 1.5 x 12 = 18 W
4. Consider the following circuit with three identical light bulbs. If bulb B burns out, what will be the brightness of bulb A and C compared to before?
a) Bulb B and C will be brighter than before.
b) Bulb A will be the same, and C will be completely off.
c) Bulb A will be dimmer, and C will be the same as before.
d) Bulb A will be the same, and C will be dimmer than before.
Answer: b) In the given circuit, bulb A is in parallel with bulbs B and C, while bulbs B and C are in series. When bulb B burns out, the circuit becomes an open circuit in the series branch containing bulbs B and C.
Since bulb A is in parallel with the series branch of bulbs B and C, it will continue to receive the same voltage as before. Therefore, bulb A will remain at the same brightness as before. Since bulb B has burned out and there is no current flowing through the series branch, bulb C will also be off, and its brightness will be zero.
5. A copper wire has a resistance of 3 Ω. If a new copper wire is taken that has the same length as the first wire but has four times the cross-sectional area, what will be the resistance of the new wire?
a) 3 Ω
b) 12 Ω
c) 0.75 Ω
d) 0.18 Ω
Answer: c) The resistance of a wire is indeed inversely proportional to its cross-sectional area, assuming the length and resistivity remain constant. The formula for resistance is:
R = (ρ x L) /A
Given that the first copper wire has a resistance (R_{1}) of 3 ohms and the new copper wire has four times the cross-sectional area, we can compare their resistances using the ratio of their cross-sectional areas:
R_{1}/R_{2} = ((ρ.L)/A_{1})/((ρ.L)/A_{2})
Since the length and resistivity are the same for both wires, they cancel out.
R_{1}/R_{2} = A_{2}/A_{1}
3/R_{2} = 4/1
R_{2} = 3/4
R_{2} = 0.75 Ω
So, the resistance of the new copper wire is 0.75 ohms.
1. How is Electricity Generated?
Burning fossil fuels, nuclear reactions, hydroelectric power (using water), wind power, solar power, and other technologies can all be used to create electricity. Every technique involves converting various forms of energy, such as solar, kinetic, nuclear, or chemical energy, into electrical energy.
2. What are Conductors and Insulators?
Conductors are materials that allow electric charges to flow through them easily, such as metals like copper and aluminium. Insulators, on the other hand, are materials that do not allow electric charges to flow freely, like rubber, plastic, and wood. Understanding the difference is crucial for safely using and managing electricity.
3. What are the Differences Between AC and DC Electricity?
Electricity provided to homes and businesses is known as alternating current (AC), in which the direction of electric charge flow is often reversed. Batteries and solar panels are popular sources of DC (Direct Current) power, which is defined as an electric charge flowing in a single direction.
4. What are Electrical Circuits?
An electrical circuit is a path or loop through which electricity flows. Circuits can be simple, like those in a flashlight, or complex, like the wiring in a house. Understanding circuits is essential for troubleshooting electrical problems and for designing electrical systems.
5. How Electricity is measured?
Voltage (volts), current (amperes or amps), and resistance (ohms) are the units used to measure electricity. Watts are used to measure power, which is the rate at which electrical energy is transformed into another kind of energy.
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