Solutions for Exercises of AC Circuits
1) If the effective current in a 50 cycle AC circuit is 5 A, what is the peak value of current? What is the current 1/600 sec after it was zero?
f = 50Hz, Irms = 5A,
I0 =?, I =?after t = 1/600 s
We know that, $latex I_{rms}=\frac{I_0}{\sqrt2}$
$latex I_0=\sqrt2I_{rms}$
∴ I0 = 1.41 × 5 = 7.07A
Instantaneous expression for current is given as,
I = I0sin(2πft)
I = 7.07sin(100πt)
Hence at t = 1/600, We have,
$latex I\;=\;7.07\sin\left(100\pi\times\frac1{600}\right)=7.07\sin\left(\frac\pi6\right)$
$latex \therefore\;I\;=\;7.07\;\times\;\frac12$
$latex \therefore\;I\;=\;3.535A$
2) A light bulb is rated 100W for 220 V AC supply of 50 Hz. Calculate (a) resistance of the bulb. (b) the rms current through the bulb.
Prated = 100W, Vrated = 220V, f = 50Hz
(a) R =? (b) Irms =?
P = VRMS × IRMS
$latex P\;=\;\frac{{V^2}_{rms}}R$
$latex R\;=\;\frac{{V^2}_{rms}}P$
$latex \therefore\;R\;=\;\frac{200^2}{100}=484\Omega$
$latex \&\;I_{RMS}=\frac P{V_{RMS}}$
$latex \;I_{RMS}=\frac{100}{220}=0.45A$
3) A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current?
C = 15μF = 15 × 10-6F, Vrms = 220 V, f = 50 Hz,
XC =?, Irms =? I0 =?
XC =?Irms =? for f′ = 2f
$latex X_C\;=\;\frac1{2\pi fC}$
$latex =\;\frac1{2\times3.14\times50\times15\times10^{-6}}$
$latex \therefore\;X_C=\;212\Omega$
$latex I_{rms}\;=\frac{V_{rms}}{X_C}$
$latex I_{rms}\;=1.037A$
$latex I_0\;=\sqrt2I_{rms}$
$latex I_0\;=1.41\;\times\;1.037A$
$latex I_0\;=\;1.47A$
$latex X_C=\frac1{2\pi fC}\;\;\;\;\;\;\;\;\therefore X_C\;\propto\;\frac1f$
Hence capacitive reactance will be halved if the frequency is doubled.
$latex I_{rms}=\;\frac{V_{rms}}{X_C}$
$latex \therefore I_{rms}\;\propto\;\frac1{X_C}\propto\;f$
Hence Irms is doubled if the frequency is doubled.
4) An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor (π = 3.142)
L = 2H, I0 = 0.25A, f = 60Hz, Vrms =?
$latex I_{rms}\;=\frac1{\sqrt2}$
$latex \therefore\;I_{rms}\;=\frac{0.25}{1.41}$
$latex \therefore\;I_{rms}\;=0.178A$
Vrms = Irms × XL = Irms × 2πfL
∴ Vrms = 0.178 × 2 × 3.141 × 60 × 2
∴ Vrms = 133.5V
5) Alternating emf of e = 220 sin 100t is applied to a circuit containing an inductance of (1/) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?
e = e0sin(100πt), L=1/πH,
Instantaneous Current & AC galvanometer reading = ?
Comparing e = e0sin(100πt) with standard expression for instantaneous emf, e = e0sin(2πft)
we have, e0 = 220V and 2πf = 100π ∴ f = 50Hz
$latex X_L\;=\;2\pi fL\;=\;2\;\times\;\pi\;\times\;50\;\times\;\frac1\pi$
$latex X_L\;=\;100\Omega$
Instantaneous expression for current is given as,
$latex i\;=\;i_0\sin\left(2\pi ft-\frac\pi2\right)$
$latex i\;=\;\left(\frac{e_0}{X_L}\right)\sin\left(2\pi ft-\frac\pi2\right)$
$latex \therefore\;i\;=\;\left(\frac{220}{100}\right)\sin\left(100\pi t-\frac\pi2\right)$
$latex \therefore\;i\;=\;2.2\sin\;\left(100\pi t-\frac\pi2\right)$
AC galvanometer reads rms value, hence we have,
$latex I_{rms}=\frac{I_0}{\sqrt2}=\frac{2.2}{1.41}=1.56A$
6) A 25 µF capacitor, a 0.10 H inductor and a 25Ω resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?
C = 25μF = 25 × 10-6F, L = 0.1H, R = 25Ω,
e = 310sin(314t)V; f = ?, X = ?, Z = ?, I = ?, ∅ = ?
Given emf is e = 310sin(314t) comparing this expression with e = e0sin(2πft), we have,
e0 = 310V and 2πf = 314
$latex \therefore\;f\;=\;\frac{314}{2\times3.14}=50Hz$
XL = 2πfL
XL = 2 × ???? × 50 ×0.1
XL = 31.41Ω
$latex X_C=\frac1{2\pi fC}$
$latex X_C=\frac1{2\times3.14\times50\times25\times10^{-6}}$
$latex X_C=127.32\Omega$
Hence effective reactance is,
X = XC − XL
∴ X = 127.32 − 31.41
∴ X = 95.90Ω
$latex Z\;=\;\sqrt{X^2\;+\;R^2\;}$
$latex Z\;=\;\sqrt{\left(95.92\right)^2\;+\;25^2\;}$
$latex Z\;=\;99.10\Omega$
$latex I_{rms}=\frac{e_{rms}}Z=\frac{\left({\displaystyle\frac{e_0}{\sqrt2}}\right)}Z$
$latex i.e.\;I_{rms}=\frac{310}{1.41\times99.10}=2.21A$
$latex \cos\phi=\frac RZ$
$latex \cos\phi=\frac{25}{99.10}$
$latex \cos\phi=0.253$
$latex \therefore\;\phi\;=\;\cos^{-1}(0.253\;)\;=\;1.31rad\;$
7) A capacitor of 100 µF, a coil of resistance 50Ω and an inductance 0.5 H are connected in series with a 110 V-50Hz source. Calculate the rms value of current in the circuit.
C = 100μF = 10-4F, R = 50Ω, , L = 0.5H, f = 50 Hz
Vrms = 110V, Vrms =?
$latex Z\;=\;\sqrt{R^2\;+\;{(X_L\sim X_C)}^2\;}$
XL = 2πfL
= 2 × ???? × 50 × 0.5
XL = 157.07Ω
$latex X_C=\frac1{2\pi fC}$
$latex X_C=\frac1{2\times\pi\times50\times10^{-4}}$
XC = 31.8Ω
$latex \therefore\;Z\;=\sqrt{{(50)}^2\;+\;{(157.07\;-\;31.8)}^2}$
$latex \therefore\;Z\;=134.87\Omega$
$latex \therefore\;I_{rms}=\frac{V_{rms}}Z$
$latex \therefore\;I_{rms}=\frac{110}{134.87}$
$latex \therefore\;I_{rms}=0.816A$
8) Find the capacity of a capacitor which when put in series with a 10Ω resistor makes the power factor equal to 0.5. Assume an 80V-100Hz AC supply.
R = 10Ω, Vrms = 80V, f = 100 Hz,
C = ? for power factor = 0.5
$latex i.e\;X_C=\frac1{2\pi fC}$
$latex \therefore\;X_C=\frac1{2\times\pi\times100\times C}$
$latex \therefore\;X_C=\frac1{628C}$
$latex \therefore\;Z\;=\;\sqrt{R^2\;+\;{(X_C)}^2\;}$
$latex \therefore\;Z\;=\;\sqrt{10^2\;+\;\left(\frac1{628C}\right)^2\;}$
∴ Power factor = R/Z
$latex =\frac{10}{\sqrt{10^2+\left({\displaystyle\frac1{628C}}\right)}^2}=0.5$
$latex \therefore\;10\;=\;0.5\sqrt{10^2+\left(\frac1{628C}\right)^2}$
$latex \therefore\;20\;=\;\sqrt{10^{2\;}+\;\left(\frac1{628C}\right)^2}$
$latex \therefore\;400\;=\;100\;+\;\left(\frac1{628C}\right)^2$
$latex \therefore\;\left(\frac1{628C}\right)^2\;=\;300$
$latex \therefore\;\frac1{628C}\;=\;17.32$
$latex \therefore\;C\;=\;9.2\;\times\;10^{-5}F$
9) Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
f = 50 Hz, time t = ? to reach I = Irms
We know that,$latex I\;=\;I_0\sin(2\pi ft)\;\&$
$latex I_{rm}\;=\;\frac{I_0}{\sqrt2}$
Hence for I = Irms, we have,
$latex \frac{I_0}{\sqrt2}=I_0\sin\left(2\pi ft\right)$
$latex \therefore\frac1{\sqrt2}=\sin\left(2\pi ft\right)$
$latex \therefore\;2\pi ft\;=\frac\pi4\;as\;\sin^{-1}\left(\frac1{\sqrt2}\right)=\frac\pi4$
$latex \therefore\;t\;=\;\frac1{8\times50}$
t = 2.5 × 10-3s
10) Calculate the value of capacity in picofarad, which will make 101.4 micro henry inductance to oscillate with frequency of one megahertz.
L = 101.4μH = 101.4 × 10-6H,
fr = 1MHz = 106Hz, C = ?
$latex f_r=\frac1{2\pi\sqrt{LC}}$
$latex i.e\;C=\frac1{4\pi^2Lf_r^2}$
$latex \therefore C=\frac1{4\times\left(3.14\right)^2\times101.4\times10^{-6}\times10^{12}}$
$latex \therefore C=2.498\;\times\;10^{-1}F$
$latex \therefore\;C\;=\;249.8\;\times\;10^{-12}F$
$latex \therefore\;C=\;249.8pF\;$
11) A 10 µF capacitor is charged to a 25 volt of potential. The battery is disconnected and a pure 100 mH coil is connected across the capacitor so that LC oscillations are set up. Calculate the maximum current in the coil.
C = 10μF = 10 × 10-6F , V = 25 V,
L = 100mH = 10-1H, Imax =?
When capacitor is fully charged energy stored is,
$latex i.e\;E=\frac12CV^2$
$latex \therefore\;E=\frac12\times10\times10^{-6}\times25^2$
$latex \therefore\;E=312.5\times10^{-5}J$
This electrostatics energy is converted into magnetic energy within the inductor. But maximum energy stored in the inductor is given as,
$latex E\;=\;\frac12L{I^2}_{max}$
$latex \therefore=\;\frac12L{I^2}_{max}=312.5\times10^{-5}$
$latex {I^2}_{max}=\frac{2\times312.5\times10^{-5}}{10^{-1}}$
$latex {I^2}_{max}=625\times10^{-4}$
∴ Imax = 25 × 10-2
∴ Imax = 0.25A
12) A 100µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.
C = 100μF = 100 × 10-6F, V = 50 V,
Imax = 5A, L = ?
When capacitor is fully charged energy stored is,
$latex E=\frac12CV^2$
$latex E=\frac12\;\times\;100\;\times\;10^{-6}\;\times\;50^2$
$latex E=0.125\;J$
This electrostatics energy is converted into magnetic energy within the inductor. But maximum energy stored in the inductor is given as,
$latex E=\frac12L{I^2}_{max}$
$latex \therefore\frac12L{I^2}_{max}=0.125$
$latex \therefore\;L\;=\;\frac{0.25}{25}=0.01\;H$