StudentAidUnit 16: Alternating Current — Numericals
12th Class Physics · Unit 16: Alternating Current
16.1.An A.C. voltmeter reads 250 V. What is its peak and instantaneous values if the frequency of alternating voltage is 50 Hz?
Given
RMS voltage
V_{rms = 250 V
Frequency
f = 50 Hz
Formula
Peak voltage formula
V_{rms = frac{V0}{sqrt{2}
Rearranged
V0 = sqrt{2} × V_{rms
Working
Substituting
V0 = sqrt{2} × 250
= 353.5 V
Formula
Instantaneous voltage
V = V0 sin(2pi ft)
Working
Substituting
V = 353.5 sin(2pi × 50 × t)
= 353.5 sin(100pi t) V
V0 = 353.5 V; quad V = 353.5 sin(100pi t) V
16.2.A 100 μF capacitor is connected to an alternating voltage of 24 V and frequency 50 Hz. Calculate: (a) The reactance of the capacitor, and (b) The current in the circuit
Given
Capacitance
C = 100 μF = 100 × 10-6 F
RMS voltage
V_{rms = 24 V
Frequency
f = 50 Hz
Formula
Capacitive reactance
XC = 12pi fC
Working
Substituting values for X_C
XC = frac{1}{2 × 3.14 × 50 × 100 × 10-6
XC = 31.8 Ω
Formula
Current in capacitive circuit
I_{rms = frac{V_{rms{XC}
Working
Substituting
I_{rms = 2431.8
= 0.75 A
XC = 31.8 Ω; quad I_{rms = 0.75 A
16.3.When 10 V are applied to an A.C. circuit, the current flowing in it is 100 mA. Find its impedance.
Given
RMS voltage
V_{rms = 10 V
RMS current
I_{rms = 100 mA = 100 × 10-3 A
Formula
Impedance definition
Z = frac{V_{rms{I_{rms
Working
Substituting
Z = frac{10}{100 × 10-3
= 100 Ω
Z = 100 Ω
16.4.At what frequency will an inductor of 1.0 H have a reactance of 500 Ω?
Given
Inductance
L = 1.0 H
Inductive reactance
XL = 500 Ω
Formula
Inductive reactance formula
XL = omega L = 2pi fL
Solving for frequency
f = XL2pi L
Working
Substituting
f = 5002 × 3.14 × 1.0
= 5006.28 = 80 Hz
f = 80 Hz
16.5.An iron core coil of 2.0 H and 50 Ω is placed in series with a resistance of 450 Ω. An A.C. supply of 100 V, 50 Hz is connected across the circuit. Find (i) the current flowing in the coil, (ii) phase angle between the current and voltage.
Given
Coil resistance
R1 = 50 Ω
Series resistance
R2 = 450 Ω
Inductance
L = 2.0 H
RMS voltage
V_{rms = 100 V
Frequency
f = 50 Hz
Formula
Inductive reactance
omega L = 2pi fL = 2 × 3.14 × 50 × 2.0
= 628 Ω
Total resistance
R = R1 + R2 = 50 + 450
= 500 Ω
Impedance
Z = sqrt{R2 + (omega L)2}
Working
Substituting
Z = sqrt{(500)2 + (628)2}
= 803 Ω
Formula
Current (part i)
I_{rms = frac{V_{rms{Z}
Working
Substituting
I_{rms = 100803
= 0.01245 A = 12.45 mA
Formula
Phase angle (part ii)
θ = tan-1left(omega LRright)
Working
Substituting
θ = tan-1left(628500right)
= 51.5^circ
I_{rms = 12.45 mA; quad θ = 51.5^circ
16.6.A circuit consists of a capacitor of 2 μF and a resistance of 1000 Ω connected in series. An alternating voltage of 12 V and frequency 50 Hz is applied. Find (i) the current in the circuit, and (ii) the average power supplied.
Given
Resistance
R = 1000 Ω
Capacitance
C = 2 μF = 2 × 10-6 F
RMS voltage
V_{rms = 12 V
Frequency
f = 50 Hz
Formula
Capacitive reactance
XC = 12pi fC
Working
Substituting
XC = frac{1}{2 × 3.14 × 50 × 2 × 10-6
= 1592 Ω
Formula
Impedance
Z = sqrt{R2 + (XC)2}
Working
Substituting
Z = sqrt{(1000)2 + (1592)2}
= 1880 Ω
Formula
Current (part i)
I_{rms = frac{V_{rms{Z}
Working
Substituting
I_{rms = 121880
= 0.0064 A = 6.4 mA
Formula
Phase angle
θ = tan-1left(XCRright)
Working
Substituting
θ = tan-1left(15921000right)
= 57.87^circ
Formula
Average power (part ii)
P = V_{rms I_{rms cos θ
Working
Substituting
P = 12 × 0.0064 × cos 57.87^circ
= 0.04 W
I_{rms = 6.4 mA; quad P = 0.04 W
16.7.Find the capacitance required to construct a resonance circuit of frequency 1000 kHz with an inductor of 5 mH.
Given
Frequency
f = 1000 kHz = 106 Hz
Inductance
L = 5 mH = 5 × 10-3 H
Formula
Resonance frequency formula
f = frac{1}{2pisqrt{LC}
Squaring both sides
f2 = 14pi2 LC
Solving for C
C = 14pi2 L f2
Working
Substituting values
C = 14(3.14)2 × 5 × 10-3 × (106)2
= 5.09 PF
C = 5.09 PF