9th Grade Alternating Current Concepts: Cycle, Period, Frequency, Capacitors, Inductors, Power, and Work Practice Questions & Solutions

Solved Example

Easy Level

💡 An alternating current (AC) signal completes 60 cycles every second.

Question: What is the frequency of this AC signal?
What is its period?
Describe what one cycle of this AC signal represents.

Solution & Explanation

Here's how to solve this problem step-by-step:

👉 Understanding Frequency:
Frequency is defined as the number of cycles per second.
In this case, the signal completes 60 cycles in 1 second.
Therefore, the frequency (\(f\)) is 60 cycles/second, which is 60 Hertz (Hz).
\[ f = 60 \text{ Hz} \]
👉 Understanding Period:
The period (\(T\)) is the time it takes for one complete cycle. It is the reciprocal of the frequency.
\[ T = \frac{1}{f} \]
Substitute the frequency:
\[ T = \frac{1}{60 \text{ Hz}} \]
\[ T \approx 0.0167 \text{ seconds} \] So, the period of the AC signal is approximately 0.0167 seconds.
👉 Understanding a Cycle:
A cycle of this AC signal represents one complete back-and-forth oscillation or variation of the voltage or current. For example, if we start at zero voltage, the voltage would rise to a positive peak, decrease back to zero, then go to a negative peak, and finally return to zero. This entire sequence is one cycle.

✅ Summary:
The frequency is 60 Hz.
The period is approximately 0.0167 seconds.
One cycle is a complete variation of the AC signal from its starting point, through its positive and negative peaks, and back to the starting point.

Solved Example

Easy Level

📌 A specific AC power outlet provides voltage that completes one full cycle in 0.02 seconds.

Question: What is the period of this AC voltage?
Calculate its frequency.
If this outlet is in a country using a common standard, which country might it be (e.g., USA or Europe)?

Solution & Explanation

Let's break down the solution:

👉 Identifying the Period:
The problem states that the voltage completes one full cycle in 0.02 seconds. By definition, the time for one complete cycle is the period (\(T\)).
\[ T = 0.02 \text{ seconds} \]
👉 Calculating the Frequency:
Frequency (\(f\)) is the reciprocal of the period (\(T\)).
\[ f = \frac{1}{T} \]
Substitute the given period:
\[ f = \frac{1}{0.02 \text{ s}} \]
\[ f = 50 \text{ Hz} \] So, the frequency of this AC voltage is 50 Hertz.
👉 Relating to Real-World Standards:
Many countries, particularly in Europe, Asia, and Africa, use a 50 Hz AC power standard. The USA and Canada, for example, use a 60 Hz standard.
Therefore, this outlet might be in a country like Germany, the UK, India, or Australia, which use 50 Hz AC power.

✅ Summary:
The period is 0.02 seconds.
The frequency is 50 Hz.
This outlet might be in a country using the 50 Hz standard, such as many in Europe.

Solved Example

Medium Level

💡 Imagine you have a simple circuit with a light bulb and a power source. You first connect it to a Direct Current (DC) battery, then to an Alternating Current (AC) source.

Question: What is a capacitor, and what is its primary function in a circuit?
How would a capacitor behave if placed in series with the light bulb for DC current versus AC current? Explain the difference.

Solution & Explanation

Let's explore capacitors:

👉 What is a Capacitor?
A capacitor is an electrical component that can store electrical energy in an electric field. It typically consists of two conductive plates separated by an insulating material (dielectric). Think of it like a tiny, temporary battery that can quickly charge and discharge. Its primary function is to store charge and block direct current while allowing alternating current to pass (under certain conditions).
👉 Capacitor Behavior with DC Current:
If a capacitor is placed in series with the light bulb and connected to a DC battery:
- Initially, when the circuit is closed, the capacitor will start to charge up. During this charging phase, current will flow through the circuit, and the light bulb might briefly light up.
- Once the capacitor is fully charged, it acts like an open circuit (a break in the wire). It will block the flow of direct current.
- Therefore, after a very short time, the light bulb will turn off and stay off because no continuous DC current can flow through the series circuit.
👉 Capacitor Behavior with AC Current:
If the same capacitor is placed in series with the light bulb and connected to an AC source:
- With AC, the voltage and current are constantly changing direction. The capacitor will continuously charge and discharge as the AC voltage alternates.
- When the AC voltage is changing, the capacitor allows current to flow "through" it (it's not truly flowing through the dielectric, but charging and discharging creates a current flow in the circuit).
- Therefore, the light bulb will light up and stay lit (assuming the capacitor's value is appropriate for the frequency and voltage) because the AC current can effectively pass through the capacitor.

✅ Summary:
A capacitor stores energy. It blocks DC current once charged, causing the light bulb to turn off. It allows AC current to pass, keeping the light bulb lit.

Solved Example

Medium Level

📌 Consider another circuit with a light bulb and a power source. We will again test it with a DC battery and an AC source.

Question: What is an inductor, and what is its primary function in a circuit?
How would an inductor behave if placed in series with the light bulb for DC current versus AC current? Explain the difference.

Solution & Explanation

Let's understand inductors:

👉 What is an Inductor?
An inductor is an electrical component, usually a coil of wire, that can store electrical energy in a magnetic field when current flows through it. Its primary function is to oppose changes in current. It allows direct current to pass easily but resists the flow of alternating current.
👉 Inductor Behavior with DC Current:
If an inductor is placed in series with the light bulb and connected to a DC battery:
- When the circuit is closed, the inductor initially resists the sudden change in current (due to its property called inductance). This might cause the light bulb to light up gradually.
- Once the current becomes steady (DC), the inductor essentially acts like a short circuit or a simple wire (assuming ideal conditions, with very low resistance in the wire coil).
- Therefore, the light bulb will light up and stay lit brightly because the DC current flows easily through the inductor.
👉 Inductor Behavior with AC Current:
If the same inductor is placed in series with the light bulb and connected to an AC source:
- With AC, the current is constantly changing direction and magnitude. An inductor's nature is to oppose these changes in current.
- This opposition to changing AC current is called inductive reactance. The faster the current changes (higher frequency), the more the inductor opposes the flow.
- Therefore, the inductor will resist the flow of AC current. The light bulb will either glow dimly or not light up at all, depending on the frequency of the AC and the inductor's value. It effectively "blocks" or significantly reduces AC current.

✅ Summary:
An inductor stores energy in a magnetic field and opposes changes in current. It allows DC current to pass easily, making the light bulb bright. It resists AC current, making the light bulb dim or off.

Solved Example

Medium Level

💡 A common household appliance, like a hair dryer, operates on an AC voltage of 120 V and draws a current of 10 A.

Question: Calculate the power consumed by the hair dryer.
If the hair dryer is used for 15 minutes, how much work (energy) does it do in Joules?

Solution & Explanation

Let's calculate the power and work:

👉 Calculating Power:
For a purely resistive AC appliance like a hair dryer, the average electrical power (\(P\)) consumed can be calculated using the formula:
\[ P = V \times I \] Where:
- \(V\) = Voltage = 120 V
- \(I\) = Current = 10 A
Substitute the values:
\[ P = 120 \text{ V} \times 10 \text{ A} \]
\[ P = 1200 \text{ Watts (W)} \] So, the power consumed by the hair dryer is 1200 Watts.
👉 Calculating Work (Energy):
Work done or energy consumed (\(W\)) is the product of power and time (\(t\)).
\[ W = P \times t \] First, convert the time from minutes to seconds, as the standard unit for energy (Joule) requires time in seconds.
\[ t = 15 \text{ minutes} \times 60 \text{ seconds/minute} = 900 \text{ seconds} \] Now, substitute the power and time into the formula:
\[ W = 1200 \text{ W} \times 900 \text{ s} \]
\[ W = 1,080,000 \text{ Joules (J)} \] So, the work (energy) done by the hair dryer in 15 minutes is 1,080,000 Joules.

✅ Summary:
The power consumed is 1200 W.
The energy (work) done is 1,080,000 J.

Solved Example

Medium Level

📌 An LED television has a power rating of 80 W. It is connected to a standard AC outlet.

Question: If the television is left on for 5 hours continuously, how much electrical energy (work) does it consume in kilowatt-hours (kWh)?
If the cost of electricity is 0.15 per kWh, what is the total cost of running the TV for 5 hours?

Solution & Explanation

Let's calculate the energy consumption and cost:

👉 Calculating Electrical Energy (Work) in kWh:
Energy consumed (\(W\)) is calculated as Power (\(P\)) multiplied by Time (\(t\)).
\[ W = P \times t \] We need the energy in kilowatt-hours (kWh). First, convert the power from Watts to kilowatts (kW) and time from minutes to hours.
- Power (\(P\)) = 80 W
- To convert Watts to kilowatts: \(80 \text{ W} \div 1000 = 0.08 \text{ kW}\)
- Time (\(t\)) = 5 hours
Now, substitute these values:
\[ W = 0.08 \text{ kW} \times 5 \text{ hours} \]
\[ W = 0.4 \text{ kWh} \] So, the television consumes 0.4 kilowatt-hours of electrical energy.
👉 Calculating the Total Cost:
The cost is calculated by multiplying the total energy consumed by the cost per kilowatt-hour.
Cost = Energy consumed (kWh) \(\times\) Cost per kWh
Cost = \(0.4 \text{ kWh} \times \0.15/\text{kWh}\)
Cost = \(\0.06\) So, the total cost of running the TV for 5 hours is 0.06 (or 6 cents).

✅ Summary:
The TV consumes 0.4 kWh of energy.
The total cost to run the TV for 5 hours is 0.06.

Solved Example

Medium Level

💡 Imagine you are an engineer designing a new power distribution system for a city. You have two main options for transmitting electricity from the power plant to homes: Direct Current (DC) or Alternating Current (AC).

Question: Explain why Alternating Current (AC) is predominantly used for long-distance power transmission and distribution to homes, rather than Direct Current (DC). Focus on the key advantages of AC relevant to its properties.

Solution & Explanation

Here's why AC is favored for power distribution:

👉 Ease of Voltage Transformation:
The most significant advantage of AC is its ability to easily change voltage levels using a device called a transformer.
- For long-distance transmission, electricity is stepped up to very high voltages (e.g., hundreds of thousands of volts). This reduces the current for a given power, which in turn significantly reduces energy loss as heat (\(P_{loss} = I^2R\)) in the transmission lines.
- Before reaching homes and businesses, the high voltage AC can be easily stepped down to safer, usable voltages (e.g., 120 V or 240 V) using other transformers.
- DC voltage, on the other hand, is very difficult and inefficient to step up or down without complex and expensive electronic converters.
👉 Generation Efficiency:
AC is naturally generated by rotating generators (alternators) in power plants, which are mechanically simpler and more efficient to build and operate than DC generators for large-scale production.
👉 Safety and Switching:
AC currents can be interrupted (switched on/off) more easily and safely than high-voltage DC currents. DC arcs (electrical discharges) are much harder to extinguish, posing safety challenges for switches and circuit breakers.

✅ Summary:
AC is used for power distribution primarily because its voltage can be easily stepped up and down using transformers. This allows for efficient long-distance transmission at high voltages (minimizing energy loss) and safe distribution to consumers at lower voltages.

Solved Example

Real World Example

💡 You're traveling from the United States (which uses 60 Hz AC) to a country in Europe (which typically uses 50 Hz AC). You bring an electric toothbrush that is designed to work with 60 Hz AC.

Question: What is the primary difference between a 50 Hz and a 60 Hz AC power supply in terms of cycle and period?
What might happen if you plug your 60 Hz electric toothbrush into a 50 Hz outlet without a proper converter? Explain why.

Solution & Explanation

Let's look at the real-world implications of AC frequency:

👉 Difference between 50 Hz and 60 Hz:
The primary difference lies in their frequency (number of cycles per second) and consequently their period (time per cycle).
- 60 Hz AC: Completes 60 cycles every second. Its period is \(T = \frac{1}{60} \approx 0.0167\) seconds.
- 50 Hz AC: Completes 50 cycles every second. Its period is \(T = \frac{1}{50} = 0.02\) seconds.
This means that a 50 Hz current changes direction (alternates) 50 times per second, while a 60 Hz current changes direction 60 times per second. The 50 Hz current is "slower" in its oscillation.
👉 Impact on the Electric Toothbrush:
Many appliances, especially those with motors or timing circuits (like your electric toothbrush), are designed for a specific frequency.
If you plug a 60 Hz electric toothbrush into a 50 Hz outlet:
- Motor Speed: Electric motors rely on the AC frequency to determine their speed. A motor designed for 60 Hz will likely run slower when supplied with 50 Hz power. This is because the magnetic fields driving the motor are changing direction less frequently.
- Overheating/Damage: Running a motor at an unintended frequency can cause it to draw more current than designed, potentially leading to overheating and damage to the motor or the appliance's internal electronics. The motor might struggle, vibrate excessively, or simply not work effectively.
- Reduced Performance: The toothbrush might not vibrate or spin as powerfully, making it less effective at cleaning.
- Not all appliances are affected equally: Simple resistive loads (like a basic heater or incandescent light bulb) are generally less affected by frequency differences (though slight power output changes might occur). However, devices with motors, clocks, or sophisticated power supplies are very sensitive to frequency.
To safely use the toothbrush, you would need a voltage and frequency converter, not just a simple plug adapter.

✅ Summary:
50 Hz and 60 Hz differ in how quickly the current alternates. Plugging a 60 Hz toothbrush into a 50 Hz outlet can cause its motor to run slower, overheat, or even be damaged due to the mismatched operating frequency.

9th Grade Other: Alternating Current Concepts: Cycle, Period, Frequency, Capacitors, Inductors, Power, and Work Practice Questions

Example 1:

Solution:

Here's how to solve this problem step-by-step:

👉 Understanding Frequency:
Frequency is defined as the number of cycles per second.
In this case, the signal completes 60 cycles in 1 second.
Therefore, the frequency (\(f\)) is 60 cycles/second, which is 60 Hertz (Hz).
\[ f = 60 \text{ Hz} \]
👉 Understanding Period:
The period (\(T\)) is the time it takes for one complete cycle. It is the reciprocal of the frequency.
\[ T = \frac{1}{f} \]
Substitute the frequency:
\[ T = \frac{1}{60 \text{ Hz}} \]
\[ T \approx 0.0167 \text{ seconds} \] So, the period of the AC signal is approximately 0.0167 seconds.
👉 Understanding a Cycle:
A cycle of this AC signal represents one complete back-and-forth oscillation or variation of the voltage or current. For example, if we start at zero voltage, the voltage would rise to a positive peak, decrease back to zero, then go to a negative peak, and finally return to zero. This entire sequence is one cycle.

Example 2:

Solution:

Let's break down the solution:

👉 Identifying the Period:
The problem states that the voltage completes one full cycle in 0.02 seconds. By definition, the time for one complete cycle is the period (\(T\)).
\[ T = 0.02 \text{ seconds} \]
👉 Calculating the Frequency:
Frequency (\(f\)) is the reciprocal of the period (\(T\)).
\[ f = \frac{1}{T} \]
Substitute the given period:
\[ f = \frac{1}{0.02 \text{ s}} \]
\[ f = 50 \text{ Hz} \] So, the frequency of this AC voltage is 50 Hertz.
👉 Relating to Real-World Standards:
Many countries, particularly in Europe, Asia, and Africa, use a 50 Hz AC power standard. The USA and Canada, for example, use a 60 Hz standard.
Therefore, this outlet might be in a country like Germany, the UK, India, or Australia, which use 50 Hz AC power.

✅ Summary:
The period is 0.02 seconds.
The frequency is 50 Hz.
This outlet might be in a country using the 50 Hz standard, such as many in Europe.

Example 3:

Solution:

Let's explore capacitors:

👉 What is a Capacitor?
A capacitor is an electrical component that can store electrical energy in an electric field. It typically consists of two conductive plates separated by an insulating material (dielectric). Think of it like a tiny, temporary battery that can quickly charge and discharge. Its primary function is to store charge and block direct current while allowing alternating current to pass (under certain conditions).
👉 Capacitor Behavior with DC Current:
If a capacitor is placed in series with the light bulb and connected to a DC battery:
- Initially, when the circuit is closed, the capacitor will start to charge up. During this charging phase, current will flow through the circuit, and the light bulb might briefly light up.
- Once the capacitor is fully charged, it acts like an open circuit (a break in the wire). It will block the flow of direct current.
- Therefore, after a very short time, the light bulb will turn off and stay off because no continuous DC current can flow through the series circuit.
👉 Capacitor Behavior with AC Current:
If the same capacitor is placed in series with the light bulb and connected to an AC source:
- With AC, the voltage and current are constantly changing direction. The capacitor will continuously charge and discharge as the AC voltage alternates.
- When the AC voltage is changing, the capacitor allows current to flow "through" it (it's not truly flowing through the dielectric, but charging and discharging creates a current flow in the circuit).
- Therefore, the light bulb will light up and stay lit (assuming the capacitor's value is appropriate for the frequency and voltage) because the AC current can effectively pass through the capacitor.

✅ Summary:
A capacitor stores energy. It blocks DC current once charged, causing the light bulb to turn off. It allows AC current to pass, keeping the light bulb lit.

Example 4:

Solution:

Let's understand inductors:

👉 What is an Inductor?
An inductor is an electrical component, usually a coil of wire, that can store electrical energy in a magnetic field when current flows through it. Its primary function is to oppose changes in current. It allows direct current to pass easily but resists the flow of alternating current.
👉 Inductor Behavior with DC Current:
If an inductor is placed in series with the light bulb and connected to a DC battery:
- When the circuit is closed, the inductor initially resists the sudden change in current (due to its property called inductance). This might cause the light bulb to light up gradually.
- Once the current becomes steady (DC), the inductor essentially acts like a short circuit or a simple wire (assuming ideal conditions, with very low resistance in the wire coil).
- Therefore, the light bulb will light up and stay lit brightly because the DC current flows easily through the inductor.
👉 Inductor Behavior with AC Current:
If the same inductor is placed in series with the light bulb and connected to an AC source:
- With AC, the current is constantly changing direction and magnitude. An inductor's nature is to oppose these changes in current.
- This opposition to changing AC current is called inductive reactance. The faster the current changes (higher frequency), the more the inductor opposes the flow.
- Therefore, the inductor will resist the flow of AC current. The light bulb will either glow dimly or not light up at all, depending on the frequency of the AC and the inductor's value. It effectively "blocks" or significantly reduces AC current.

Example 5:

Solution:

Let's calculate the power and work:

👉 Calculating Power:
For a purely resistive AC appliance like a hair dryer, the average electrical power (\(P\)) consumed can be calculated using the formula:
\[ P = V \times I \] Where:
- \(V\) = Voltage = 120 V
- \(I\) = Current = 10 A
Substitute the values:
\[ P = 120 \text{ V} \times 10 \text{ A} \]
\[ P = 1200 \text{ Watts (W)} \] So, the power consumed by the hair dryer is 1200 Watts.
👉 Calculating Work (Energy):
Work done or energy consumed (\(W\)) is the product of power and time (\(t\)).
\[ W = P \times t \] First, convert the time from minutes to seconds, as the standard unit for energy (Joule) requires time in seconds.
\[ t = 15 \text{ minutes} \times 60 \text{ seconds/minute} = 900 \text{ seconds} \] Now, substitute the power and time into the formula:
\[ W = 1200 \text{ W} \times 900 \text{ s} \]
\[ W = 1,080,000 \text{ Joules (J)} \] So, the work (energy) done by the hair dryer in 15 minutes is 1,080,000 Joules.

✅ Summary:
The power consumed is 1200 W.
The energy (work) done is 1,080,000 J.

Example 6:

Solution:

Let's calculate the energy consumption and cost:

👉 Calculating Electrical Energy (Work) in kWh:
Energy consumed (\(W\)) is calculated as Power (\(P\)) multiplied by Time (\(t\)).
\[ W = P \times t \] We need the energy in kilowatt-hours (kWh). First, convert the power from Watts to kilowatts (kW) and time from minutes to hours.
- Power (\(P\)) = 80 W
- To convert Watts to kilowatts: \(80 \text{ W} \div 1000 = 0.08 \text{ kW}\)
- Time (\(t\)) = 5 hours
Now, substitute these values:
\[ W = 0.08 \text{ kW} \times 5 \text{ hours} \]
\[ W = 0.4 \text{ kWh} \] So, the television consumes 0.4 kilowatt-hours of electrical energy.
👉 Calculating the Total Cost:
The cost is calculated by multiplying the total energy consumed by the cost per kilowatt-hour.
Cost = Energy consumed (kWh) \(\times\) Cost per kWh
Cost = \(0.4 \text{ kWh} \times \0.15/\text{kWh}\)
Cost = \(\0.06\) So, the total cost of running the TV for 5 hours is 0.06 (or 6 cents).

✅ Summary:
The TV consumes 0.4 kWh of energy.
The total cost to run the TV for 5 hours is 0.06.

Example 7:

Solution:

Here's why AC is favored for power distribution:

👉 Ease of Voltage Transformation:
The most significant advantage of AC is its ability to easily change voltage levels using a device called a transformer.
- For long-distance transmission, electricity is stepped up to very high voltages (e.g., hundreds of thousands of volts). This reduces the current for a given power, which in turn significantly reduces energy loss as heat (\(P_{loss} = I^2R\)) in the transmission lines.
- Before reaching homes and businesses, the high voltage AC can be easily stepped down to safer, usable voltages (e.g., 120 V or 240 V) using other transformers.
- DC voltage, on the other hand, is very difficult and inefficient to step up or down without complex and expensive electronic converters.
👉 Generation Efficiency:
AC is naturally generated by rotating generators (alternators) in power plants, which are mechanically simpler and more efficient to build and operate than DC generators for large-scale production.
👉 Safety and Switching:
AC currents can be interrupted (switched on/off) more easily and safely than high-voltage DC currents. DC arcs (electrical discharges) are much harder to extinguish, posing safety challenges for switches and circuit breakers.

Example 8:

Solution:

Let's look at the real-world implications of AC frequency:

👉 Difference between 50 Hz and 60 Hz:
The primary difference lies in their frequency (number of cycles per second) and consequently their period (time per cycle).
- 60 Hz AC: Completes 60 cycles every second. Its period is \(T = \frac{1}{60} \approx 0.0167\) seconds.
- 50 Hz AC: Completes 50 cycles every second. Its period is \(T = \frac{1}{50} = 0.02\) seconds.
This means that a 50 Hz current changes direction (alternates) 50 times per second, while a 60 Hz current changes direction 60 times per second. The 50 Hz current is "slower" in its oscillation.
👉 Impact on the Electric Toothbrush:
Many appliances, especially those with motors or timing circuits (like your electric toothbrush), are designed for a specific frequency.
If you plug a 60 Hz electric toothbrush into a 50 Hz outlet:
- Motor Speed: Electric motors rely on the AC frequency to determine their speed. A motor designed for 60 Hz will likely run slower when supplied with 50 Hz power. This is because the magnetic fields driving the motor are changing direction less frequently.
- Overheating/Damage: Running a motor at an unintended frequency can cause it to draw more current than designed, potentially leading to overheating and damage to the motor or the appliance's internal electronics. The motor might struggle, vibrate excessively, or simply not work effectively.
- Reduced Performance: The toothbrush might not vibrate or spin as powerfully, making it less effective at cleaning.
- Not all appliances are affected equally: Simple resistive loads (like a basic heater or incandescent light bulb) are generally less affected by frequency differences (though slight power output changes might occur). However, devices with motors, clocks, or sophisticated power supplies are very sensitive to frequency.
To safely use the toothbrush, you would need a voltage and frequency converter, not just a simple plug adapter.

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💡 9th Grade Other: Alternating Current Concepts: Cycle, Period, Frequency, Capacitors, Inductors, Power, and Work Practice Questions

9th Grade Other: Alternating Current Concepts: Cycle, Period, Frequency, Capacitors, Inductors, Power, and Work Practice Questions

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