Alternating current part I

 

 

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Capacitor and Inductor Circuit

Oscillation Circuit
When charging a capacitor, the capacitor gains stored electrical energy due to the electric field \[𝑈_𝐸 = \frac{{{1}}}{{{2}}} \frac{{{q^2}}}{{{C}}}\] and as a result of current passing through an inductor, it stores magnetic energy due to the magnetic field \[P = ∆𝑉_{𝑖𝑛𝑑} . i = L\frac{{{di}}}{{{dt}}}i \] \[𝑈_B =\int P . dt= \int L .i di= \frac{{{1}}}{{{2}}} L .i^2 \] Question: How does an oscillation circuit work?

Physics Exam: Capacitor and Inductor Circuit (Oscillation Circuit)

When charging a capacitor, the capacitor gains stored electrical energy due to the electric field. As a result of current passing through an inductor, it stores magnetic energy due to the magnetic field.

Question 1: What type of energy does a capacitor store when charged?

a) Kinetic energy

b) Electrical energy

c) Thermal energy

d) Chemical energy

Question 2: What causes energy storage in an inductor?

a) Electric field

b) Magnetic field

c) Electrical resistance

d) Electrical voltage

Question 3: What is the circuit called that contains a capacitor and inductor?

a) Series circuit

b) Oscillation circuit

c) Resistance circuit

d) Direct current circuit

Question 4: What happens to energy in an ideal oscillation circuit?

a) Converts from electrical to magnetic and vice versa

b) Dissipates as heat

c) Remains only electrical energy

d) Remains only magnetic energy

Question 5: What factor determines the oscillation frequency in the circuit?

a) Capacitance and inductance value

b) Battery voltage only

c) Wire resistance

d) Wire length

Question 6: When energy in the capacitor is at maximum, what is it in the inductor?

a) Also at maximum

b) At minimum

c) Zero

d) Undefined

Question 7: What happens if we add resistance to an oscillation circuit?

a) Energy dissipates over time

b) Oscillation frequency increases

c) Oscillation stops immediately

d) Oscillation is unaffected

Question 8: When current in the circuit is at maximum, where is most energy concentrated?

a) In the inductor

b) In the capacitor

c) Equal between capacitor and inductor

d) No energy in the circuit

Question 9: When voltage across the capacitor is at maximum in a circuit, what is the current in the circuit?

a) Zero

b) At maximum

c) Half of maximum

d) Changes randomly

Question 10: In what practical application are oscillation circuits used?

a) Radio devices

b) Electric motors

c) Batteries

d) Electric lights

AC Circuits

Useful Information: Operation of AC Circuits

Rotation of a coil in a magnetic field leads to the generation of induced emf given by the relation:

\[∆𝑉_{emf} =N. A.B.W Sin ( Wt) \]

AC Generator

\[𝑉_{max} =N. A.B.W \]

\[∆𝑉_{emf} =𝑉_m Sin ( Wt) \]

Thus, an induced current is produced that doesn't necessarily match the voltage in phase and varies in value every moment and direction every half cycle

\[I_{t} =I_m Sin ( Wt - ∅ ) \]

(∅) Phase angle between voltage and current

(I max) Maximum value of AC current called the amplitude of AC current

AC Circuit Containing Ohmic Resistance

In this simulation, an AC current is connected to an ohmic resistor and the voltage and current in the circuit are monitored

Note that the current and voltage measurements change value every moment The voltage and current will be represented as vectors

Experiment Results

Impedance Calculation	Voltage and Current Equations	Phase Difference Between Current and Voltage
\[R = \frac{{{𝑉 (max)}}}{{{I (max)}}}\]\[R =\frac{{{𝑉(t)}}}{{{I(t)}}}\]	Voltage Equation \[V_R =𝑉_m Sin ( Wt) \]	Graph of voltage and current over time
No relationship between frequency change and resistance	\[I_{t} =I_m Sin ( Wt) \]	Current and voltage are in phase

Example 1

An AC circuit connected to a time-varying emf source Voltage equation of the source \[V(t) = 210 Sin (60 𝜋 t ) \]The source is connected to an ohmic resistor of \[R=20 Ω \]Then the AC current equation representing this circuit is:

Click here to show solution method

Choose the correct answer

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𝐼(t) = 10.5 Sin (60𝜋 t+ 𝜋/2 )

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𝐼(t) = 14.7 Sin (60𝜋 t -𝜋/2)

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𝐼(t) = 10.5 Sin (60𝜋 t )

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𝐼(t) = 14.7 Sin (60𝜋 t +𝜋/2)

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Short Exam: AC Circuit with Ohmic Resistance

1. If the voltage equation in an AC circuit with ohmic resistance is
v(t) = 100 sin(314t),
what is the current frequency?

a) 50 Hz
b) 60 Hz
c) 314 Hz
d) 100 Hz

Solution Method:

Voltage equation: v(t) = V_m sin(ωt)

From equation: ω = 314 rad/s

Frequency f = ω / 2π = 314 / (2 × 3.14) ≈ 50 Hz

Correct answer: a) 50 Hz

2. In an AC circuit containing only ohmic resistance, the phase angle between voltage and current is:

a) 90 degrees
b) 0 degrees
c) 45 degrees
d) 180 degrees

Solution Method:

In ohmic resistance, voltage and current are in phase (no phase difference)

Correct answer: b) 0 degrees

3. How does increasing frequency affect the ohmic resistance value in an AC circuit?

a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Solution Method:

Ohmic resistance doesn't depend on frequency in AC circuits

Correct answer: c) Remains constant

4. If the current equation in an AC circuit with ohmic resistance is
i(t) = 5 sin(120πt),
what is the voltage at
t = 1/240 S
if
R = 10Ω?

a) 50 volts
b) 25 volts
c) 0 volts
d) 5 volts

Solution Method:

1. Calculate current at t = 1/240 second:

i(1/240) = 5 sin(120π × 1/240) = 5 sin(π/2) = 5 × 1 = 5A

2. Calculate voltage: v = i × R = 5A × 10Ω = 50V

Correct answer: a) 50V

5. Which of the following represents the correct relationship between voltage and current over time in an AC circuit with ohmic resistance?

a) Voltage and current are identical
b) Voltage lags current by 90 degrees
c) Voltage leads current by 90 degrees
d) Voltage and current are in phase but may have different values

Solution Method:

In ohmic resistance, voltage and current are in phase but may have different values depending on resistance value

Correct answer: d) Voltage and current are in phase but may have different values

6. If the voltage equation is
v(t) = 150 sin(377t + 30°)
across a 15 ohm resistor, what is the current equation?

a) i(t) = 10 sin(377t + 30°)
b) i(t) = 10 sin(377t)
c) i(t) = 150 sin(377t + 30°)
d) i(t) = 22.5 sin(377t + 30°)

Solution Method:

1. Calculate current amplitude: I_m = V_m / R = 150 / 15 = 10A

2. In ohmic resistance, current and voltage are in phase

3. Current equation: i(t) = 10 sin(377t + 30°)

Correct answer: a) i(t) = 10 sin(377t + 30°)

7. If voltage and current in an AC circuit with ohmic resistance are as shown below (two sine waves identical in phase but different in amplitude), what is the resistance value if voltage amplitude is 200 volts and current amplitude is 4 amps?

a) 50 ohms
b) 800 ohms
c) 0.02 ohms
d) 25 ohms

Solution Method:

Resistance = Voltage amplitude / Current amplitude

R = V_m / I_m = 200V / 4A = 50Ω

Correct answer: a) 50Ω

AC Circuit Containing Inductive Coil (Neglecting Ohmic Resistance)

In this simulation, an AC current is connected to an inductive coil and the voltage and current in the circuit are monitored

Note that the current and voltage measurements change value every moment The voltage and current will be represented as vectors

Experiment Results

Inductive Reactance Calculation	Voltage and Current Equations	Phase Difference Between Current and Voltage
\[XL = \frac{{{𝑉 (max)}}}{{{I (max)}}}\]\[XL =\frac{{{𝑉(t)}}}{{{I(t)}}}\]\[XL = 2𝜋 f .L \]	Voltage Equation \[V_{t} =𝑉_{mL} Sin ( Wt) \]	Graph of voltage and current over time
Relationship between frequency and inductive reactance Direct relationship \[XL\propto f\] Slope \[m=\frac{XL}{f} =2𝜋 L\]	Current Equation\[v = L\frac{di}{dt}=𝑉_m Sin ( Wt) \]\[dI=\frac{𝑉_m}{L}Sin ( Wt)dt\]\[I= \int \frac{𝑉_m}{L}Sin ( Wt)dt\]\[I=-\frac{𝑉_m}{LW}Cos ( Wt)\]\[I=-\frac{𝑉_m}{XL}Cos ( Wt)\]\[I=- I_mCos ( Wt)\]\[-Cos ( Wt)=Sin ( Wt-\frac{𝜋}{2})\]\[I= I_m Sin ( Wt-\frac{𝜋}{2})\]	Voltage leads current by 90 degrees

Example 2

An AC circuit connected to a time-varying emf source Voltage equation of the source
\[V(t) = 210 Sin (60 𝜋 t ) \]
The source is connected to a coil with self-inductance \[L=0.2 H \]
Then the AC current equation representing this circuit is:

Click here to show solution method

Choose the correct answer

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𝐼(t) = 5.67 Sin (60𝜋 t + 𝜋/2 )

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𝐼(t) = 17.5 Sin (60𝜋 t + 𝜋/2 )

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𝐼(t) = 5.57 Sin (60𝜋 t - 𝜋/2 )

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𝐼(t) = 17.5 Sin (60𝜋 t - 𝜋/2 )

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AC Circuit Containing Capacitive Reactance

In this simulation, an AC current is connected to a capacitive reactance and the voltage and current in the circuit are monitored

Note that the current and voltage measurements change value every moment The voltage and current will be represented as vectors

Experiment Results

Capacitive Reactance Calculation	Voltage and Current Equations	Phase Difference Between Current and Voltage
\[XC = \frac{{{𝑉 (max)}}}{{{I (max)}}}\]\[XC =\frac{{{𝑉(t)}}}{{{I(t)}}}\]\[Xc = \frac{{{1}}}{{{2𝜋 f .C}}}\]	Voltage Equation \[v_R =𝑉_m Sin ( Wt) \]	Graph of voltage and current over time
Relationship between frequency and capacitive reactance Inverse relationship \[XC\propto \frac{1}{f}\] Slope \[m=XCf =\frac{1}{2𝜋.C}\]	Current Equation\[q = C .V = C .𝑉_m Sin ( Wt) \]\[i=\frac{dq}{dt}\]\[i= \frac{d(C .𝑉_m Sin ( Wt))}{dt}\]\[i=W.C.{𝑉_m}Cos ( Wt)\]\[XC=\frac{1}{WC}\]\[i=\frac{𝑉_m}{XC}Cos ( Wt)=I_mCos ( Wt)\]\[Cos ( Wt)=Sin ( Wt+\frac{𝜋}{2})\]\[i= I_m Sin ( Wt+\frac{𝜋}{2})\]	Current leads voltage by 90 degrees

Example 3

(C = = 40 𝜇F ) A capacitive reactance with capacitance
connected to an AC source Voltage equation of the source
\[V_c =210 Sin (60𝜋 t) \]
Then the AC current equation representing this circuit is:

Click here to show solution method

Choose the correct answer

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𝑖 (C ) =1.58 𝑆𝑖𝑛 (60𝜋 𝑡 + 𝜋/2)

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𝑖 (C ) =1.58 𝑆𝑖𝑛 (60𝜋 𝑡 - 𝜋/2)

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𝑖 (C ) =2.25 𝑆𝑖𝑛 (60𝜋 𝑡 + 𝜋/2)

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𝑖 (C ) =2.25 𝑆𝑖𝑛 (60𝜋 𝑡 - 𝜋/2)

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