# Ch. 24 HW Solutions

advertisement
```Ch. 24 HW Solutions
2
7.) The average power lost in
√ a resistor is given by P = I R where I is the rms value
of current, given by I = io / 2, and R is the resistance of the resistor. The current in
the √
problem has amplitude io = 5.0 A and the resistor has R = 15 Ω. So P = I 2 R =
(io / 2)2 R = i2o R/2 = 187.5W .
9.) (a.) A V = 80 V AC power source and a resistor in series yield a current I = V /R =
80/30 = 2.67 A. (b.) An 80 V power source has an rms voltage
of 80 V, though
the amplitude
√
√
of the voltage oscillations will be greater by a factor of 2. √
(c.) vo = 2V = 113.14V (d.)
With an rms current of I = 2.67 A, the peak current is io = 2I = 3.77 A.
13.) For discharging a capacitor, q(t) = qo e(−t/τC ) . (a.) After four time constants, t = 4τC so
q(4τ ) = qo e(−4τC /τC ) = qo e−4 = qo (0.37)4 = (0.0187)qo . The fraction is 0.0187. (b.) Initially,
qo = CVo . Finally, qf = CVf . Dividing the latter by the former, qf /qo = (CVf /CVo ) = Vf /Vo .
Since qf /qo = 0.0187 = Vf /Vo , then Vf = (0.0187)Vo .
21.) (a.) An AC circuit with an inductor abides by V = IXL where V and I are the
rms voltage and current and XL is the impedence of the inductor. With V and I given,
XL = V /I = 10000 Ω. (b.) The impedance of a coil is given by XL = 2πf L. With the
impedence and frequency konwn, L = XL /(2πf ) = 1.592 H.
23.) L and f given, XL = 2πf L. (a.) I = V /XL = 1.31 A. (b.) For f = 6000 Hz,
I = V /XL = 0.0131 A.
27.) An AC, RL circuit has L = 3 × 10−3 p
H, R = 200 Ω, V = 50 V, and f = 10, 000
Hz. (a.) XL = 2πf L = 188.5 Ω. (b.) ZRL = R2 + XL2 = 274.8Ω. (c.) I = V /ZRL = 0.182
A. (d.) VR = IR = 36.39 V. (e.) VL = IXL = 36.29 V.
33.) The air-core solenoid’s ohmmeter reading gives the resistance of the wire in the coil
(NOT the impedence!). Placed in series with the capacitor, the two effectively √
create an
RLC circuit. Given C and the resonant frequency, fo , and using fo = 1/(2π LC) we
solve: L = 1/(4π 2 fo2 C) = 5.87 × 10−5 H. To find the voltage across the inductor we
must first find the current in the circuit. At resonance the impedence of the capacitor
and inductor are equal and thus cancel, giving Z = R. So the current in the circuit is
I = V /R = 6.30 A. To find the voltage across the inductor we must include both the
inducedpemf from thepinductor and the voltage drop from the resistance in the coil. So
ZRL = R2 + XL2 = R2 + (2πfo L)2 = 6.046 Ω. VL = VRL = IZRL = 38.07 V.
37.) The current and the voltages across eachp
circuit element in an RCL circuit are given.
(a.) The power factor is given by cos φ = VR / VR2 + (VL − VC )2p= 0.707. (b.) The source
voltage is equal to the voltage across all circuit elements, or V = VR2 + (VL − VC )2 = 42.43
V. (c.) The average power output of the source is P = IV cos φ or, since power is only
dissipated in the resistor, P = IVR = 90 W. The energy in the inductor and capacitor is not
disspated; it is transferred back and forth between them.
√
47.) Given R, L, and C, we calculate fo = 1/(2π LC) = 5032.9 Hz. At resonance, the
impedence is just Z = R = 100 Ω.
51.) A reversal of current direction indicates we go from 6 A to -6 A in 0.01 seconds,
∆i
= 12/0.01 = 1200 A/s. The solenoid is to be 10 cm in diameter so it has an area of
so ∆t
A = π(d/2)2 = 0.00785 m2 . To induce a 200 V potential in a 0.4 m long solenoid, we note
∆i
. So the number of turns
L = µo n2 Al = µo (N 2 /l2 )Al = µo N 2 A/l. Then εind = µo N 2 A/l ∆t
q
εind l
needed is N = µ A ∆i = 2599 turns ≈ 2600 turns.
o
∆t
2
```
##### Related flashcards
Symbols

28 Cards

Electric power

25 Cards

Signal processing

22 Cards

Computer hardware

24 Cards