Alternating Current Numerical Problems - Chapter 16 of FSc/ICS Physics

Alternating Current Numerical Problems - Chapter 16

Problem 1: An alternating current is represented by the equation \( I = 20 \sin(100\pi t) \). Compute its frequency and the maximum and the rms value of current.

Given Data:

• Equation of current \( I = 20 \sin(100\pi t) \)

To Determine:

• Frequency \( f = ? \)
• Maximum Current \( I_0 = ? \)
• RMS Current \( I_{RMS} = ? \)

Calculations:

General Equation of A.C. \( I = I_0 \sin(2\pi ft) \)

Comparing with given equation \( I = 20 \sin(100\pi t) \):

\[ 2\pi ft = 100\pi t \Rightarrow f = \frac{100\pi}{2\pi} = 50 \, \text{Hz} \]

\[ I_0 = 20 \, \text{A} \]

\[ I_{RMS} = \frac{I_0}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 14.14 \, \text{A} \]

Problem 2: A solenoidal A.C. has a maximum value of 15 A. What is its rms values? If the time is recorded from the instant the current is zero and is becoming positive, what is the instantaneous value of current after 1/300 s, given the frequency is 50 Hz.

Given Data:

• Maximum Current \( I_0 = 15 \, \text{A} \)
• Time \( t = \frac{1}{300} \, \text{s} \)
• Frequency \( f = 50 \, \text{Hz} \)

To Determine:

• RMS Current \( I_{RMS} = ? \)
• Instantaneous Current \( I = ? \)

Calculations:

\[ I_{RMS} = \frac{I_0}{\sqrt{2}} = \frac{15}{\sqrt{2}} = 10.61 \, \text{A} \]

Equation of Instantaneous Alternating Current:

\[ I = I_0 \sin(2\pi ft) \]

\[ = 15 \sin\left(2\pi \times 50 \times \frac{1}{300}\right) \]

\[ = 15 \sin\left(\frac{100\pi}{300}\right) = 15 \sin\left(\frac{\pi}{3}\right) \]

\[ = 15 \times \frac{\sqrt{3}}{2} = 12.99 \, \text{A} \]

Problem 3: Find the values of the current and inductive reactance when A.C. voltage of 220 V at 50 Hz is passed through an inductor of 10 H.

Given Data:

• Voltage \( V = 220 \, \text{V} \)
• Frequency \( f = 50 \, \text{Hz} \)
• Inductance \( L = 10 \, \text{H} \)

To Determine:

• Inductive Reactance \( X_L = ? \)
• Current \( I = ? \)

Calculations:

\[ X_L = \omega L = 2\pi fL \]

\[ = 2 \times 3.14 \times 50 \times 10 = 3140 \, \Omega \]

\[ I = \frac{V}{X_L} = \frac{220}{3140} = 0.07 \, \text{A} \]

Problem 4: An inductor has an inductance of 1/π H and resistance of 2000 Ω. A 50 Hz A.C. is supplied to it. Calculate the reactance and impedance offered by the circuit.

Given Data:

• Inductance \( L = \frac{1}{\pi} \, \text{H} \)
• Resistance \( R = 2000 \, \Omega \)
• Frequency \( f = 50 \, \text{Hz} \)

To Determine:

• Inductive Reactance \( X_L = ? \)
• Impedance \( Z = ? \)

Calculations:

\[ X_L = \omega L = 2\pi fL \]

\[ = 2\pi \times 50 \times \frac{1}{\pi} = 100 \, \Omega \]

\[ Z = \sqrt{R^2 + X_L^2} \]

\[ = \sqrt{(2000)^2 + (100)^2} \]

\[ = \sqrt{4,000,000 + 10,000} = \sqrt{4,010,000} \]

\[ = 2002.5 \, \Omega \]

Problem 5: An inductor of pure inductance 3/π H is connected in series with a resistance of 40 Ω. Find (i) the peak value of current (ii) the rms value, and (iii) the phase difference between the current and the applied voltage \( V = 350 \sin(100 \pi t) \).

Given Data:

• Inductance \( L = \frac{3}{\pi} \, \text{H} \)
• Resistance \( R = 40 \, \Omega \)
• Voltage Equation \( V = 350 \sin(100 \pi t) \)

To Determine:

• Peak Value of Current \( I_0 = ? \)
• RMS Current \( I_{RMS} = ? \)
• Phase Difference \( \varphi = ? \)

Calculations:

From voltage equation: \( V_0 = 350 \, \text{V} \), \( 2\pi f = 100\pi \Rightarrow f = 50 \, \text{Hz} \)

\[ X_L = \omega L = 2\pi fL = 2\pi \times 50 \times \frac{3}{\pi} = 300 \, \Omega \]

\[ Z = \sqrt{R^2 + X_L^2} = \sqrt{(40)^2 + (300)^2} = \sqrt{1600 + 90000} = \sqrt{91600} = 302.66 \, \Omega \]

\[ I_0 = \frac{V_0}{Z} = \frac{350}{302.66} = 1.156 \, \text{A} \]

\[ I_{RMS} = \frac{I_0}{\sqrt{2}} = \frac{1.156}{\sqrt{2}} = 0.817 \, \text{A} \]

\[ \varphi = \tan^{-1} \left( \frac{X_L}{R} \right) = \tan^{-1} \left( \frac{300}{40} \right) = \tan^{-1}(7.5) = 82.4^\circ \]

Problem 6: A 10 mH, 20 Ω coil is connected across 240 V and 180/π Hz source. How much power does it dissipate?

Given Data:

• Inductance \( L = 10 \, \text{mH} = 0.01 \, \text{H} \)
• Resistance \( R = 20 \, \Omega \)
• Voltage \( V = 240 \, \text{V} \)
• Frequency \( f = \frac{180}{\pi} \, \text{Hz} \)

To Determine:

• Power Dissipation \( P = ? \)

Calculations:

\[ X_L = \omega L = 2\pi fL = 2\pi \times \frac{180}{\pi} \times 0.01 = 3.6 \, \Omega \]

\[ Z = \sqrt{R^2 + X_L^2} = \sqrt{(20)^2 + (3.6)^2} = \sqrt{400 + 12.96} = \sqrt{412.96} = 20.32 \, \Omega \]

\[ I_{RMS} = \frac{V}{Z} = \frac{240}{20.32} = 11.81 \, \text{A} \]

\[ \cos\varphi = \frac{R}{Z} = \frac{20}{20.32} = 0.984 \]

\[ P = V I_{RMS} \cos\varphi = 240 \times 11.81 \times 0.984 = 2789.6 \, \text{W} \]

Problem 7: Find the value of the current flowing through a capacitance 0.5 μF when connected to a source of 150 V at 50 Hz.

Given Data:

• Capacitance \( C = 0.5 \, \mu\text{F} = 0.5 \times 10^{-6} \, \text{F} \)
• Voltage \( V = 150 \, \text{V} \)
• Frequency \( f = 50 \, \text{Hz} \)

To Determine:

• Current \( I = ? \)

Calculations:

\[ X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \]

\[ = \frac{1}{2 \times 3.14 \times 50 \times 0.5 \times 10^{-6}} \]

\[ = \frac{1}{157 \times 10^{-6}} = 6369.4 \, \Omega \]

\[ I = \frac{V}{X_C} = \frac{150}{6369.4} = 0.0235 \, \text{A} \]

Problem 8: An alternating source of emf 12 V and frequency 50 Hz is applied to a capacitor of capacitance 3 μF in series with a resistor of resistance of 1 kΩ. Calculate the phase angle.

Given Data:

• EMF \( E = 12 \, \text{V} \)
• Frequency \( f = 50 \, \text{Hz} \)
• Capacitance \( C = 3 \, \mu\text{F} = 3 \times 10^{-6} \, \text{F} \)
• Resistance \( R = 1 \, \text{k}\Omega = 1000 \, \Omega \)

To Determine:

• Phase Angle \( \varphi = ? \)

Calculations:

\[ X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \]

\[ = \frac{1}{2 \times 3.14 \times 50 \times 3 \times 10^{-6}} \]

\[ = \frac{1}{942 \times 10^{-6}} = 1061.6 \, \Omega \]

\[ \varphi = \tan^{-1} \left( \frac{X_C}{R} \right) = \tan^{-1} \left( \frac{1061.6}{1000} \right) = \tan^{-1}(1.0616) = 46.7^\circ \]

Problem 9: What is the resonance frequency of a circuit which include a coil of inductance 2.5 H and a capacitance 40 μF?

Given Data:

• Inductance \( L = 2.5 \, \text{H} \)
• Capacitance \( C = 40 \, \mu\text{F} = 40 \times 10^{-6} \, \text{F} \)

To Determine:

• Resonant Frequency \( f_R = ? \)

Calculations:

\[ f_R = \frac{1}{2\pi\sqrt{LC}} \]

\[ = \frac{1}{2 \times 3.14 \times \sqrt{2.5 \times 40 \times 10^{-6}}} \]

\[ = \frac{1}{6.28 \times \sqrt{100 \times 10^{-6}}} = \frac{1}{6.28 \times 10^{-2}} \]

\[ = \frac{1}{0.0628} = 15.92 \, \text{Hz} \]

Problem 10: An inductor of inductance 150 μH is connected in parallel with a variable capacitor whose capacitance can be changed from 500 pF to 20 pF. Calculate the maximum frequency and minimum frequency for which the circuit can be tuned.

Given Data:

• Inductance \( L = 150 \, \mu\text{H} = 150 \times 10^{-6} \, \text{H} \)
• Minimum Capacitance \( C_1 = 500 \, \text{pF} = 500 \times 10^{-12} \, \text{F} \)
• Maximum Capacitance \( C_2 = 20 \, \text{pF} = 20 \times 10^{-12} \, \text{F} \)

To Determine:

• Minimum Frequency \( f_1 = ? \)
• Maximum Frequency \( f_2 = ? \)

Calculations:

Minimum Frequency (with maximum capacitance):

\[ f_1 = \frac{1}{2\pi\sqrt{LC_1}} \]

\[ = \frac{1}{2 \times 3.14 \times \sqrt{150 \times 10^{-6} \times 500 \times 10^{-12}}} \]

\[ = \frac{1}{6.28 \times \sqrt{7.5 \times 10^{-14}}} = \frac{1}{6.28 \times 2.738 \times 10^{-7}} \]

\[ = \frac{1}{1.72 \times 10^{-6}} = 0.581 \times 10^6 \, \text{Hz} = 0.581 \, \text{MHz} \]

Maximum Frequency (with minimum capacitance):

\[ f_2 = \frac{1}{2\pi\sqrt{LC_2}} \]

\[ = \frac{1}{2 \times 3.14 \times \sqrt{150 \times 10^{-6} \times 20 \times 10^{-12}}} \]

\[ = \frac{1}{6.28 \times \sqrt{3 \times 10^{-15}}} = \frac{1}{6.28 \times 5.477 \times 10^{-8}} \]

\[ = \frac{1}{3.44 \times 10^{-7}} = 2.91 \times 10^6 \, \text{Hz} = 2.91 \, \text{MHz} \]