Delmar's Standard Textbook Of Electricity
7th Edition
ISBN: 9781337900348
Author: Stephen L. Herman
Publisher: Cengage Learning
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Chapter 23, Problem 1RQ
What is the phase angle relationship of current and the voltage dropped across a pure resistance?
Expert Solution & Answer
To determine
The phase angle relationship of current and voltage dropped across a pure resistance.
Answer to Problem 1RQ
The phase angle between current and voltage in a pure resistance is zero degrees.
Explanation of Solution
In a pure resistance the current flowing in the resistance is in phase with the voltage drop across it. We say that phase difference between both parameters is zero degrees. The same can be seen in the phasor diagram shown above.
The ohms law for pure resistive circuit is written as,
It can be seen that there is no presence of any imaginary term in the equation above. Hence the phase angle is zero degrees.
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PSD
A certain signal f(t) has the following PSD (assume 12 load):
| Sƒ(w) = π[e¯\w\ + 8(w − 2) + +8(w + 2)]
(a) What is the mean power in the bandwidth w≤ 1 rad/sec?
(b) What is the mean power in the bandwidth 0.99 to 1.01 rad/sec?
(c) What is the mean power in the bandwidth 1.99 to 2.01 rad/sec?
(d) What is the total mean power in (t)?
Pav=
+
2T
SfLw) dw
- SALW)
An AM modulation waveform signal:-
p(t)=(8+4 cos 1000πt + 4 cos 2000πt) cos 10000nt
(a) Sketch the amplitude spectrum of p(t).
(b) Find total power, sideband power and power efficiency.
(c) Find the average power containing of each sideband.
Can you rewrite the solution because it is
unclear?
AM
(+) = 8(1+ 0.5 cos 1000kt +0.5 ros 2000ks)
=
cos 10000 πt.
8 cos wat + 4 cos wit + 4 cos Wat coswet.
-Jet
jooort
J11000 t
= 4 e
jqooort jgoort
+4e
+ e
+e
j 12000rt.
12000 kt
+
e
+e
jooxt
igoo t
te
(w) = 8ES(W- 100007) + 8IS (W-10000)
USB
Chapter 23 Solutions
Delmar's Standard Textbook Of Electricity
Ch. 23 - What is the phase angle relationship of current...Ch. 23 - What is the phase angle relationship of current...Ch. 23 - What is the phase angle relationship of current...Ch. 23 - An AC circuit has a frequency of 400 Hz. A 16-...Ch. 23 - If 440 V are connected to the circuit, how much...Ch. 23 - Prob. 6RQCh. 23 - What is the true power of the circuit in Question...Ch. 23 - What is the apparent power of the circuit in...Ch. 23 - What is the power factor of the circuit in...Ch. 23 - How many degrees are the voltage and current out...
Ch. 23 - You are an electrician working in a plant. A...Ch. 23 - The circuit shown in Figure 23-2 is connected to a...Ch. 23 - The circuit is connected to a 400-Hz line with an...Ch. 23 - The circuit is connected to a 60-Hz line. The...Ch. 23 - This circuit is connected to a 1000-Hz line. The...Ch. 23 - A series RLC circuit contains a 4-k resistor, an...Ch. 23 - A series RLC circuit contains a resistor with a...Ch. 23 - Is the power factor in Question 6 a leading or...Ch. 23 - A series RLC circuit contains a resistor with a...Ch. 23 - A series RLC circuit has an applied voltage of 240...Ch. 23 - A series RLC circuit is connected to a 60-Hz power...
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- Can you rewrite the solution because it is unclear? AM (+) = 8(1+0.5 cos 1000kt +0.5 ros 2000 thts) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. J4000 t j11000rt $14+) = 45 jqooort +4e + e + e j 12000rt. 12000 kt + e +e +e Le jsoort -; goon t te +e Dcw> = 885(W- 100007) + 8 IS (W-10000) - USBarrow_forwardCan you rewrite the solution because it is unclear? Q2 AM ①(+) = 8 (1+0.5 cos 1000πt +0.5 ros 2000kt) $4+) = 45 = *cos 10000 πt. 8 cos wat + 4 cosat + 4 cos Wat coswet. j1000016 +4e -j10000πt j11000Rt j gooort -j 9000 πt + e +e j sooort te +e J11000 t + e te j 12000rt. -J12000 kt + с = 8th S(W- 100007) + 8 IS (W-10000) <&(w) = USB -5-5 -4-5-4 b) Pc 2² = 64 PSB = 42 + 4 2 Pt Pc+ PSB = y = Pe c) Puss = PLSB = = 32 4² = 8 w 32+ 8 = × 100% = 140 (1)³×2×2 31 = 20% x 2 = 3w 302 USB 4.5 5 5.6 6 ms Ac = 4 mi = 0.5 mz Ac = 4 ५ M2 = =0.5arrow_forwardA. Draw the waveform for the following binary sequence using Bipolar RZ, Bipolar NRZ, and Manchester code. Data sequence= (00110100) B. In a binary PCM system, the output signal-to-quantization ratio is to be hold to a minimum of 50 dB. If the message is a single tone with fm-5 kHz. Determine: 1) The number of required levels, and the corresponding output signal-to-quantizing noise ratio. 2) Minimum required system bandwidth.arrow_forward
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