Alternating currents | Class 12 NEB Physics | Numerical problems discussion
AC Circuit Problems
Physics · Alternating Current
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A circuit consists of a capacitor of 2μF and a resistor of 1000Ω. 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.
The current in this circuit is given by:
\[I=\frac{V}{\sqrt{R^2+ X_C^2}}\]First, calculate the capacitive reactance:
\[\begin{align*} X_C&=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\\ &=\frac{1}{2 \times \pi \times 50 \times 2 \times 10^{-6}}\\ &=1591.55 \hspace{0.1cm} \Omega \end{align*}\]Then the current:
\[\begin{align*} I&=\frac{12}{\sqrt{(1000)^2 + (1591.55)^2}}\\ &=0.0064 \hspace{0.1cm} \text{A} \end{align*}\]The voltage across the capacitor:
\[\begin{align*} V_C&=I\times X_C=0.0064 \times 1591.55\\ &=10.2 \hspace{0.1cm} \text{V} \end{align*}\]The phase angle between the applied emf and the current:
\[\begin{align*} \tan\phi&=\frac{X_C}{R}\\ \phi&=\tan^{-1} \frac{1591.55}{1000}\\ \therefore \phi&= 57.9^\circ \end{align*}\]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 = 200Ω, 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.
L-C-R alternating current series circuit of L = 1 H, C = 1μF and R = 100Ω are connected in series with a source of frequency 50 Hz. What is the phase shift between current and voltage?
The phase shift between current and voltage is given by:
\[\phi=\tan^{-1}\left(\frac{X_C - X_L}{R}\right)\] \[\begin{align*} X_C&=\frac{1}{2\pi f C}=\frac{1}{2 \pi \times 50 \times 10^{-6}}=3183.1 \hspace{0.1cm} \Omega\\[6pt] X_L&=2\pi f L = 2 \pi \times 50 \times 1 = 314.16 \hspace{0.1cm} \Omega \end{align*}\] \[\begin{align*} \phi&=\tan^{-1}\left(\frac{3183.1 - 314.16}{100}\right)\\ \therefore \phi&=88^\circ \end{align*}\]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.
From the impedance expressions:
\[\begin{align*} Z_1^2&=R^2+\omega_1^2 L^2 \hspace{0.1cm} \text{... (i)}\\ Z_2^2&=R^2 + \omega_2^2 L^2\hspace{0.1cm} \text{... (ii)} \end{align*}\]Subtracting (i) from (ii):
\[\begin{align*} Z_2^2-Z_1^2&=L^2(\omega_2^2 - \omega_1^2)=4\pi^2 L^2(f_2^2 - f_1^2)\\[6pt] L^2&=\frac{Z_2^2 - Z_1^2}{4\pi^2(f_2^2 - f_1^2)}=\frac{8^2-5^2}{4\pi^2(100^2-50^2)}\\ &=1.32\times 10^{-4}\\ \therefore L&=0.0114 \hspace{0.1cm} \text{H} \end{align*}\]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.
The power consumed in the L-R circuit is \(P = IV\cos\phi\). The phase angle:
\[\begin{align*} \tan\phi&=\frac{X_L}{R}=\frac{2\pi f L}{R}=\frac{2 \pi \times 50 \times 100 \times 10^{-3}}{20}\\ \therefore \phi&=\tan^{-1}(1.57)=57.5^\circ \end{align*}\]The inductive reactance and current:
\[\begin{align*} X_L&=2\pi \times 50 \times 100\times 10^{-3}=31.42 \hspace{0.1cm} \Omega\\[6pt] I&=\frac{200}{\sqrt{(20)^2+(31.42)^2}}=5.36 \hspace{0.1cm}\text{A} \end{align*}\] \[\begin{align*} P&=5.36\times 200 \times \cos 57.5°\\ \therefore P&=575.98 \hspace{0.1cm}\text{W} \end{align*}\]An iron cored coil of 2 H and 50Ω resistance placed in series with a resistor of 450Ω and 200 V, 50 Hz a.c. supply is connected across the arrangement, find
- the current flowing the coil,
- its phase angle relative to the voltage supply,
- the voltage across the coil.
The impedance of the full circuit:
\[\begin{align*} Z&=\sqrt{(R+r)^2+(X_L)^2}\\ &=\sqrt{(450+50)^2+(2\pi \times 50 \times 2)^2}\\ &=803 \hspace{0.1cm} \Omega \end{align*}\]The effective impedance of the coil alone:
\[\begin{align*} Z_L&=\sqrt{X_L^2 + r^2}=\sqrt{(628.32)^2+(50)^2}=630.31 \hspace{0.1cm} \Omega\\[6pt] V_L&=I \times Z_L = 0.25 \times 630.31 = 157.57 \hspace{0.1cm} \text{V} \end{align*}\]A 50 V a.c. supply is connected to a resistor having resistance 50Ω, 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.
Since \(I = V/Z\):
\[\begin{align*} 0.5&=\frac{50}{Z} \implies Z=100 \hspace{0.1cm} \Omega\\[6pt] \sqrt{(R+r)^2 + X_L^2}&=100\\ (50+r)^2+(78.54)^2&=(100)^2\\ (50+r)^2&=3831.47\\ 50+r&=61.89\\ \therefore r&=11.89 \hspace{0.1cm} \Omega \end{align*}\]Alternating voltage in an ac circuit is represented by V = 100√2 sin(100πt) volts. Find its root mean square value and the frequency.
\(V_0 = 100\sqrt{2}\) | \(\omega = 100\pi\)
A 100 V, 50 Hz ac source is connected to an LCR circuit containing L = 8.1 mH, C = 12.5μF and R = 10Ω all connected in series. Find the potential difference across the resistor.
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.
A circuit consists of an inductor of 200μH and resistance of 10Ω 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.
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 voltage across L is equal to that across R, calculate the voltage across R and the frequency of the supply. (Given L = 0.1 H and R = 40Ω)
Since \(V_L = V_R \implies IX_L = IR \implies X_L = R\)
\[\begin{align*} I&=\frac{V}{\sqrt{R^2+(X_L)^2}}=\frac{40}{\sqrt{R^2+R^2}}\\ &=\frac{40}{\sqrt{2\times (40)^2}}=0.71 \hspace{0.1cm} \text{A}\\[6pt] V_R&=IR=0.71 \times 40=28.4 \hspace{0.1cm} \text{V} \end{align*}\]For frequency, since \(X_L = R\):
\[\begin{align*} 2\pi f \times L &= R\\ f&=\frac{40}{2\pi \times 0.1}\\ \therefore f&=63.7 \hspace{0.1cm} \text{Hz} \end{align*}\]A constant A.C. supply is connected to a series circuit consisting of a resistance of 300Ω in series with a capacitance 6.67μF, 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.
Setting up the condition \(I_2 = \frac{1}{2}I_1\) and substituting the impedance expressions, solving for the additional resistance r:
\[\frac{V}{\sqrt{R^2+X_C^2}}=\frac{1}{2} \times \frac{V}{\sqrt{(R+r)^2+X_C^2}}\]Solving this equation yields additional resistance = 306.1Ω.
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