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howard (cah3459) – hw 23 – turner – (56705)
1
This printout should have 18 questions.
Multiplechoice questions may continue on
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beFore answering.
001 (part 1 of 2) 10.0 points
The alternating voltage oF a generator is rep
resented by the equation
E
=
E
0
sin
ω t,
where
E
is in volts,
E
0
= 230 V,
ω
= 583 rad
/
s,
and
t
is in seconds.
±ind the Frequency oF the electric potential
E
.
Correct answer: 92
.
7873 Hz.
Explanation:
Let :
E
max
=
E
0
= 230 V
,
and
ω
= 583 rad
/
s
.
ω
=2
π f
f
=
ω
2
π
=
583 rad
/
s
2
π
=
92
.
7873 Hz
.
002 (part 2 of 2) 10.0 points
±ind the voltage output oF the source.
Correct answer: 162
.
635 V.
Explanation:
The
rms
voltage is
E
rms
=
E
max
√
2
=
230 V
√
2
=
162
.
635 V
.
003 (part 1 of 2) 10.0 points
The maximum potential di²erence across cer
tain heavyduty appliances is 340 V.
The
total resistance oF an appliance is 110 Ω.
±ind the
rms
potential di²erence across the
appliance.
Correct answer: 240
.
416 V.
Explanation:
Let :
Δ
V
max
= 340 V
.
Δ
V
rms
=
√
2
2
Δ
V
max
=
Δ
V
max
√
2
=
340 V
√
2
= 240
.
416 V
004 (part 2 of 2) 10.0 points
±ind the
rms
current in the appliance.
Correct answer: 2
.
1856 A.
Explanation:
Let :
R
= 110 Ω
.
Δ
V
rms
= Δ
I
rms
R
Δ
I
rms
=
Δ
V
rms
R
=
240
.
416 V
110 Ω
=
2
.
1856 A
.
005 (part 1 of 4) 10.0 points
In the circuit shown in fgure,
E
1
=
(40 V) cos 2
πft
,
f
= 270 Hz,
E
2
= 6 V, and
R
= 40 Ω.
E
1
=
E
max
cos
ωt
E
max
= 40 V
E
2
R
±ind the maximum current through the re
sistor.
Correct answer: 1
.
15 A.
Explanation:
Let :
R
= 40 Ω
,
E
1
= (40 V) cos 2
π f t,
and
E
2
= 6 V
.
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View Full Document howard (cah3459) – hw 23 – turner – (56705)
2
Applying Kirchhof’s loop rule, we have
E
1
+
E
2

IR
=0
,
so the current is
I
(
t
)=
E
1
+
E
2
R
=
I
1
cos 2
π f t
+
I
2
,
where
I
1
=
40 V
40 Ω
= 1 A
,
and
I
2
=
6V
40 Ω
=0
.
15 A
,
since the maximum current occurs when
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

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