Alternating currents | Class 12 NEB Physics | Numerical problems discussion

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A circuit consists of a capacitor of 2μF and a resistor of 1000 Ω. An alternating emf of 12 V and frequency 50 Hz is applied. Find the voltage across the capacitor and the phase angle between the applied emf and the current.
Solution:
Given,
capacitance, C = 2μF = 2×10^-6 F
resistnce , R = 1000 Ω
alternating emf, V = 12 V
frequency, f = 50 Hz
voltage across capacitor, V_C = ?
phase angle between the applied emf and the current, φ = ?

Now, the current in this circuit is given by the expression \[I=\frac{V}{\sqrt{R^2+ X_C^2}}\] Here, \[\begin{align*} X_C&=\frac{1}{\omega C}\\ &=\frac{1}{2 \pi f C}\\ &=\frac{1}{2 \times \pi \times 50 \times 2 \times 10^{-6}}\\ &=1591.55 \hspace{0.1cm} \Omega\\ \end{align*}\] Then, \[\begin{align*} I&=\frac{V}{\sqrt{R^2 + X_C^2}}\\ &=\frac{12}{\sqrt{(1000)^2 + (1591.55)^2}}\\ &=0.0064 \hspace{0.1cm} \text{A}\\ \end{align*}\] Also, \[\begin{align*} V_C&=I\times X_C\\ &=0.0064 \times 1591.55\\ &=10.2 \hspace{0.1cm} \text{V}\\ \end{align*}\] The phase angle between the applied emf and the current is, \[\begin{align*} tan\phi&=\frac{X_C}{R}\\ \phi&=tan^{-1} \frac{1591.55}{1000}\\ \therefore \phi&= 57.9^\circ\\ \end{align*}\]
Thus, the voltage across capacitor is 10.2 V and the phase angle is 57.9⁰ .

An ac source of 220 V, 50 Hz is connected to a series circuit containing a resistor R and inductor L and a capacitor C. If R = 200 Ω , L = 0.5 H and C = 10μF, calculate, (i) the current in the circuit, (ii) the phase angle and (iii) the power consumed in the circuit.
Solution:
Given,
rms voltage, V = 220 V
frequency , f = 50 Hz
resistance, R = 200 Ω
inductance, L = 0.5 H
capacitance, C = 10 μF = 10×10^{-6} F
(i) rms current in the circuit, I = ?
(ii) phase angle, φ = ?
(iii) power consumed in the circuit, P = ?

\[\begin{align*} X_L& = \omega L\\ &= 2\pi fL\\ &=2\pi \times 50 \times 0.5\\ &=157.08 \hspace{0.1cm} \Omega\\ \end{align*}\]
\[\begin{align*} X_C&=\frac{1}{\omega C}\\ &=\frac{1}{2\pi f C}\\ &=\frac{1}{2 \pi \times 50 \times 10\times 10^{-6}}\\ &=318.31 \hspace{0.1cm} \Omega\\ \end{align*}\]
Now, \[\begin{align*} Z&=\sqrt{(X_C-X_L)^2 + R^2}\\ &=\sqrt{(318.31-157.08)^2 + (200)^2}\\ &=256.89 \hspace{0.1cm} \Omega\\ \end{align*}\] Then, (i) \[\begin{align*} I&=\frac{V}{Z}\\ &=\frac{220}{256.89}\\ &=0.856 \hspace{0.1cm} \text{A}\\ \end{align*}\]
(ii) \[\begin{align*} tan\phi&=\frac{X_C-X_L}{R}\\ &=\frac{318.31-157.08}{200}\\ \phi&=tan^{-1}(0.81)\\ \therefore \phi&=38.87^{\circ}\\ \end{align*}\]
(iii) \[\begin{align*} P&= I V cos\phi\\ &=0.856\times 220 \times cos(38.87)\\ \therefore P&=146.35 \hspace{0.1cm} \text{W}\\ \end{align*}\]
Thus, the current in the circuit is 0.856 A , phase angle is 38.87⁰ the power consumeed in the circuit is 146.35 W.

L-C-R alternating current series circuit of L = 1 H, C = 1 μF and R = 100 Ω are connected in series with a source of frequency 50 Hz. What is the phase shift between current and voltage?
Solution:
Given,
L = 1 H
C = 1 μF = 1×10^-6 F
R = 100 Ω
frequency, f = 50 Hz
phase shift, φ = ?

The phase shift between current and voltage is given by, \[\begin{align*} tan\phi&=\frac{X_C - X_L}{R}\\ \phi&=tan^{-1}\left(\frac{X_C - X_L}{R}\right)\\ \end{align*}\]
\[\begin{align*} X_C&=\frac{1}{\omega C}\\ &=\frac{1}{2\pi f C}\\ &=\frac{1}{2 \pi \times 50 \times 10^{-6}}\\ &=3183.1 \hspace{0.1cm} \Omega\\ \end{align*}\]
\[\begin{align*} X_L&=\omega L\\ &=2\pi f L\\ &=2 \pi \times 50 \times 1\\ &=100\pi\\ &=314.16 \hspace{0.1cm} \Omega\\ \end{align*}\]
So, \[\begin{align*} \phi&=tan^{-1}\left(\frac{X_C - X_L}{R}\right)\\ &=tan^{-1} \left(\frac{3183.1 - 314.16}{100}\right)\\ \therefore \phi&=88^\circ\\ \end{align*}\] Thus, the phase shift between current and voltage is 88⁰ .

A coil having inductance and resistance is connected to an oscillator giving a fixed sinusoidal output voltage of 5 V rms. With the oscillator set at a frequency of 50 Hz, the rms current in the coil is 1 A and at a frequency of 100 Hz, the rms current is 0.625 A. Determine the inductance of the coil.
Solution:
Given,
rms voltage, V = 5 V
frequency, f₁ = 50 Hz
rms current, I₁ = 1 A
frequency , f₂ = 100 Hz
rms current, I₂ = 0.625 A
inductance of the coil, L = ?

Now, \[\begin{align*} I_1&=\frac{V}{Z_1}\\ Z_1&=\frac{V}{I_1}\\ &=\frac{5}{1}\\ &=5\hspace{0.1cm} \text{A}\\ \end{align*}\]
Also, \[\begin{align*} Z_2&=\frac{V}{I_2}\\ &=\frac{5}{0.625}\\ &=8\hspace{0.1cm} \omega\\ \end{align*}\]
Then, \[\begin{align*} Z_1&=\sqrt{R^2 + (\omega_1 L)^2}\\ Z_1^2&=R^2+\omega_1^2 L^2 \hspace{0.1cm} \text{... (i)}\\ \end{align*}\]
Again, \[\begin{align*} Z_2&=\sqrt{R^2+(\omega_2 L)^2}\\ Z_2^2&=R^2 + \omega_2^2 L^2\hspace{0.1cm} \text{... (ii)}\\ \end{align*}\]
Subtracting equation (i) from (ii), we get, \[\begin{align*} Z_2^2-Z_1^2&=\omega_2^2 L^2 - \omega_1^2 L^2\\ &=L^2(\omega_2^2 - \omega_1^2)\\ &=L^2 \times 4\pi^2\times (f_2^2 - f_1^2)\\ &=4\pi^2 L^2 \times (f_2^2 - f_1^2)\\ L^2&=\frac{Z_2^2 - Z_1^2}{4\pi^2 \times (f_2^2 - f_1^2)}\\ &=\frac{8^2-5^2}{4\pi^2 \times (100^2-50^2)}\\ &=1.32\times 10^{-4}\\ \therefore L&=0.0114 \hspace{0.1cm} \text{H}\\ \end{align*}\]
Thus, the inductance of the coil is 0.0114 H.

A.C. mains of 200 volts and 50 Hz is joined to a circuit containing an inductance of 100 mH and a resistance of 20Ω in series. Calculate the power consumed.
Solution:
Given,
rms voltage, V = 200 V
frequency, f = 50 Hz
inductance, L = 100 mH = 100×10^-3 H
resistance, R = 20 Ω
power consumed, P = ?

The power consumed in the L-R circuit is given by, \[P=IV cos\phi\] The phase angle is, \[\begin{align*} tan\phi&=\frac{X_L}{R}\\ &=\frac{\omega L}{R}\\ &=\frac{2\pi f L}{R}\\ &=\frac{2 \pi \times 50 \times 100 \times 10^{-3}}{20}\\ \phi&=tan^{-1}(1.57)\\ \therefore \phi&=57.5^\circ\\ \end{align*}\]
The current in the circuit is, \[I=\frac{V}{\sqrt{R^2+X_L^2}}\]
The inductive reactance is, \[\begin{align*} X_L&=2\pi f L\\ &=2\pi \times 50 \times 100\times 10^{-3}\\ &=31.42 \hspace{0.1cm} \Omega\\ \end{align*}\]
Then, \[\begin{align*} I&=\frac{V}{\sqrt{R^2+X_L^2}}\\ &=\frac{200}{\sqrt{(20)^2+(31.42)^2}}\\ &=5.36 \hspace{0.1cm}\text{A}\\ \end{align*}\] Finally, \[\begin{align*} P&=IV cos\phi\\ &=5.36\times 200 \times cos57.5\\ \therefore P&=575.98 \hspace{0.1cm}\text{W}\\ \end{align*}\]
Thus, the power consumed in the circuit is 575.98 W.

An iron cored coil of 2 H and 50 Ω resistance placed in series with a resistor of 450 Ω and 200 V, 50 Hz a.c. supply is connected across the arrangement, find

1. the current flowing the coil,
2. its phase angle relative to the voltage supply,
3. the voltage across the coil.

Solution:
Given,
inductance, L = 2 H
internal resistance of inductor, r = 50 Ω
external resistance connected to the inductor in circuit, R = 450 Ω
rms voltage, V = 200 V
frequency, f = 50 Hz
(i) rms current, I = ?
(ii) phase angle relative to the voltage supply, \(\phi\) = ?
(iii) voltage across the coil, V_L = ?

The rms current (or simply current) in the circuit is, \[I=\frac{V}{Z}\] where, Z is the impedance of the circuit.
Calculating Z, \[\begin{align*} Z&=\sqrt{(R+r)^2+(X_L)^2}\\ &=\sqrt{(450+50)^2+(2\pi f L)^2}\\ &=802.98 \\ &=803 \hspace{0.1cm} \Omega\\ \end{align*}\] So, the current is, \[\begin{align*} I&=\frac{200}{803}\\ \therefore I&=0.25 \hspace{0.1cm} \text{A}\\ \end{align*}\]
Now, the phase angle relative to the coil can be claculated as, [\begin{align*} tan\phi&=\frac{X_L}{R+r}\\ &=\frac{2\pi f L}{500}\\ &=\frac{628.32}{500}\\ \phi&=tan^{-1}(1.26)\\ \therefore \phi&=51.5^\circ\\ \end{align*}\]
The, the voltage across the coil is, \[V_L=IZ_L\] where Z_L is the effective resistance of the inductor which is calculated as, \[\begin{align*} Z_L&=\sqrt{X_L^2 + r^2}\\ &=\sqrt{(50)^2+(628.32)^2} &=630.31 \hspace{0.1cm} \Omega\\ \end{align*}\]
So, the voltage across the coil is, \[\begin{align*} V_L&=0.25 \times 630.31\\ &=157.57 \hspace{0.1cm} \text{V}\\ \end{align*}\]
Thus, the voltage across the inductor is 157.57 V.

A 50 V a.c. supply is connected to a resistor having resistance 50 Ω, in series with a solenoid whose inductance is 0.25 H. The potential difference between the ends of the resistor is 25 V. Find the resistance of the wire of the solenoid. Take frequency of the ac source is 50 Hz.
Solution:
Given,
rms voltage, V = 50 V
resistance, R = 50 Ω
inductor, L = 0.25 H
voltage across resistor, V_R = 25 V
internal resistance of the solenoid, r = ?
frequency, f = 50 Hz

Now, \[\begin{align*} V_R&=IR\\ I&=\frac{V_R}{R}\\ &=\frac{25}{50}\\ &=0.5 \hspace{0.1cm} \text{A}\\ \end{align*}\]
Also, \[I=\frac{V}{Z}\] where, \[Z=\sqrt{(R+r)^2 + X_L^2}\] Here, \[\begin{align*} X_L&=\omega L\\ &=2\pi f L\\ &=2\pi \times 50 \times 0.25\\ &=78.54 \hspace{0.1cm} \Omega\\ \end{align*}\]
Now, \[\begin{align*} I&=\frac{V}{Z}\\ 0.5&=\frac{50}{Z}\\ Z&=\frac{50}{0.5}\\ \sqrt{(R+r)^2 + X_L^2}&=100\\ (R+r)^2+(78.54)^2&=(100)^2\\ (50+r)^2&=10000-6168.53\\ (50+r)^2&=3831.47\\ 50+r&=\sqrt{3831.47}\\ r&=61.89-50\\ \therefore r&=11.89 \hspace{0.1cm} \Omega\\ \end{align*}\]
Thus, the resistance of the wire of the solenoid is 11.89 Ω.

Alternating voltage in an ac circuit is represented by V = 100\(\sqrt{2}\) sin(100πt) volts. Find its root mean square value and the frequency.
Solution:
Given,
Comparing V = 100\(\sqrt{2}\) sin(100πt) with the standard relation V = V₀ sin(\(\omega\)t), we get,
\[V_0=100\sqrt{2}\] \[\omega = 100\pi\]
Now, \[\begin{align*} V_{rms}&=\frac{V_0}{\sqrt{2}}\\ &=\frac{100\sqrt{2}}{\sqrt{2}}\\ &=100 \hspace{0.1cm} \text{V}\\ \end{align*}\] Then, \[\begin{align*} \omega&=100\pi\\ 2\pi f&=100 \pi\\ \therefore f&=50 \hspace{0.1cm} \text{Hz}\\ \end{align*}\]
Thus, the root mean square value of voltage is 100 V and the frequency is 50 Hz.

A 100 V, 50 Hz ac source is connected to an LCR circuit containing L = 8.1 mH, C = 12.5μF and R = 10 Ω all connected in series. Find the potential difference across the resistor.
Solution:
Given,
rms voltage, V = 100 V
frequency, f = 50 Hz
L = 8.1 mH = 8.1×10^-3 H
C = 12.5 μF = 12.5×10^-6 F
R = 10 Ω
potential difference across the resistor, V_R = ?

The current in the circuit is, \[I=\frac{V}{Z}\] where, Z is the impedance.

\[\begin{align*} X_L&=\omega L\\ &=2\pi f L\\ &=2\pi \times 50 \times 8.1 \times 10^{-3}\\ &=2.54 \hspace{0.1cm} \Omega\\ \end{align*}\]
\[\begin{align*} X_C&=\frac{1}{\omega C}\\ &=\frac{1}{2\pi f C}\\ &=\frac{1}{2 \pi \times 50 \times 12.5 \times 10^{-6}}\\ &=254.65 \hspace{0.1cm} \Omega\\ \end{align*}\] Now, \[\begin{align*} I&=\frac{V}{\sqrt{R^2+(X_C-X_L)^2}}\\ &=\frac{100}{\sqrt{(10)^2+(254.65-2.54)^2}}\\ &=0.39 \hspace{0.1cm} \text{A}\\ \end{align*}\]
The potential difference across the resistor is then, \[\begin{align*} V_R&=IR\\ &=0.39 \times 10\\ \therefore V_R&=3.9 \hspace{0.1cm} \text{V}\\ \end{align*}\]
Thus, the potential difference across the resistor is 3.9 V.

An alternating voltage 10 V (rms) and 4 KHz frequency is applied to a resistor of resistance 5Ω in series with a capacitor of capacitance 10μF. Calculate the rms potential difference across the resistor and the capacitor.
Solution:
Given,
rms voltage, V = 10 V
frequency, f = 4 KHz = 4×10³ Hz
resistance, R = 5 Ω
capacitance, C = 10 μF = 10×10^-6 F
voltage across resistor, V_R = ?
voltage across capacitor, V_C = ?

Now, \[\begin{align*} X_C&=\frac{1}{\omega C}\\ &=\frac{1}{2\pi f C}\\ &=\frac{1}{2\pi \times 4\times 10^3 \times 10 \times 10^{-6}}\\ &=3.98 \hspace{0.1cm} \Omega\\ \end{align*}\]
Then the impedance is, \[\begin{align*} Z&=\sqrt{R^2+(X_C)^2}\\ &=\sqrt{(5)^2+(3.98)^2}\\ &=6.39 \hspace{0.1cm} \Omega\\ \end{align*}\] Now we calculate the current as, \[\begin{align*} I&=\frac{V}{Z}\\ &=\frac{10}{6.39}\\ &=1.56 \hspace{0.1cm} \text{A}\\ \end{align*}\]
So, the voltage across resistor is, \[\begin{align*} V_R&=IR\\ &=1.56 \times 5\\ &=7.8 \hspace{0.1cm} \text{V}\\ \end{align*}\] Also, the voltage across the capacitor is, \[\begin{align*} V_C&=IX_C\\ &=1.56 \times 3.98\\ &=6.2 \hspace{0.1cm} \text{V}\\ \end{align*}\]
Thus, the potential difference across the resistor is 7.8 V and that across the capacitor is 6.2 V.

A circuit consists of an inductor of 200 μH and resistance of 10 Ω in series with a variable capacitor and a 0.10 V (rms), 1.0 MHz supply. Calculate (i) the capacitance to give resonance, and (ii) the quality factor of the circuit at resonance.
Solution:
Given,
inductance, L = 200 μH = 200×10^-6 H
resistance, R = 10 Ω
rms voltage, V = 0.10 V
frequency, f = 1.0 MHz = 1.0\times 10^6 Hz
(i) Capacitance to give resonance, C = ?
(ii) the quality factor of the circuit at resonance, Q = ?

(i) For the resonant frequency, the capacitance should be such that \[\begin{align*} f&=\frac{1}{2\pi\sqrt{LC}}\\ 10^6&=\frac{1}{2\pi \times \sqrt{200 \times 10^{-6} \times C}}\\ \therefore C&=1.27 \times 10^{-10} \hspace{0.1cm} \text{F}\\ \end{align*}\]
(ii) Quality factor is given by, \[\begin{align*} Q&=\frac{1}{R}\times \sqrt{\frac{L}{C}}\\ &=\frac{1}{10}\times \sqrt{\frac{200\times 10^{-6}}{1.27\times 10^{-10}}}\\ &=126\\ \therefore Q&=126 \\ \end{align*}\]
Thus, the capacitance is 1.27×10^-10 F and the quality factor is 126.

A coil of inductance L and negligible resistance is in series with a resistance R. A supply voltage of 40 V(rms) is connected to them. If a voltage across L is equal to that across R, caculate the voltage across R and the frequency of the supply? (Given L = 0.1 H and R = 40 Ω)
Solution:
Given,
L = 0.1 H
R = 40 Ω
rms voltage, V = 40 V
voltage across L = voltage across R (i.e., V_L = V_R)
V_R = ?
frequency of the supply, f = ?

Since, \[\begin{align*} V_L&=V_R\\ IX_L&=IR\\ X_L&=R\\ \end{align*}\]
The current across the circuit is, \[\begin{align*} I&=\frac{V}{Z}\\ &=\frac{40}{\sqrt{R^2+(X_L)^2}}\\ &=\frac{40}{\sqrt{R^2+R^2}}\\ &=\frac{40}{\sqrt{2R^2}}\\ &=\frac{40}{\sqrt{2\times (40)^2}}\\ &=0.71 \hspace{0.1cm} \text{A}\\ \end{align*}\]
Then, \[\begin{align*} V_R&=IR\\ &=0.71 \times 40\\ &=28.4 \hspace{0.1cm} \text{V}\\ \end{align*}\] And, \[\begin{align*} X_L&=\omega L\\ 40&=2\pi f \times L \hspace{0.1cm} \because X_L=R\\ 40&=2\pi \times f \times 0.1\\ f&=\frac{40}{2\pi \times 0.1}\\ \therefore f&=63.7 \hspace{0.1cm} \text{Hz}\\ \end{align*}\] Thus, the voltage across R is 28.4 V and the frequency of the supply is 63.7 Hz.

A constant A.C. supply is connected to a series circuit consisting of a resistance of 300Ω in series with a capacitance 6.67 μF, the frequency of the supply being 3000/2π Hz. It is desired to reduce the current in the circuit to half its value. Show how this could be done by placing an additional resistance.
Solution:
Given,
reistance, R = 300 Ω
capacitance, C = 6.67 μF = 6.67×10^{-6} F
frequency of the supply, f = 3000/2π Hz
Let, the initial value of current be I₁ and the reduced value be I₂.
I₂ = \(\frac{1}{2}\) I₁
additional resistance to reduce the current be r = ?

Now, \[\begin{align*} X_C&=\frac{1}{\omega C}\\ &=\frac{1}{2\pi f C}\\ &=\frac{1}{2\pi \times \frac{3000}{2\pi}\times 6.67\times 10^{-6}}\\ &=49.98 \hspace{0.1cm} \Omega\\ \end{align*}\]
Then, \[\begin{align*} I_2&=\frac{1}{2}I_1\\ \frac{V}{R^2+(X_C)^2}&=\frac{1}{2} \times \frac{V}{(300+R_1)^2+(X_C)^2}\\ \end{align*}\] Put the value and solve the equation to get 306.1 Ω .

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