119. Current in an RL Circuit The equation governing the amount of current I (in amperes) after time t (in seconds) in a single RL circuit consisting of a resistance R (in ohms), an inductance L (in henrys), and an electromotive force E (in voles) is I = E R [ 1 − e − ( R / L ) t ] (a) If E = 120 volts, R = 10 ohms, and L = 5 henrys, how much current I 1 is flowing after 0.3 second? After 0.5 second? After 1 second? (b) What is the maximum current? (c) Graph this function I = I 1 (t), measuring I along the y -axis and t along the x -axis . (d) If E = 120 volts, R = 5 ohms, and L = 10 henrys, how much current I 2 is flowing after 0.3 second? After 0.5 second? After 1 second? (e) What is the maximum current? (f) Graph the function I = I 2 ( t ) on the same coordinate axes as I 1 ( t ) .
119. Current in an RL Circuit The equation governing the amount of current I (in amperes) after time t (in seconds) in a single RL circuit consisting of a resistance R (in ohms), an inductance L (in henrys), and an electromotive force E (in voles) is I = E R [ 1 − e − ( R / L ) t ] (a) If E = 120 volts, R = 10 ohms, and L = 5 henrys, how much current I 1 is flowing after 0.3 second? After 0.5 second? After 1 second? (b) What is the maximum current? (c) Graph this function I = I 1 (t), measuring I along the y -axis and t along the x -axis . (d) If E = 120 volts, R = 5 ohms, and L = 10 henrys, how much current I 2 is flowing after 0.3 second? After 0.5 second? After 1 second? (e) What is the maximum current? (f) Graph the function I = I 2 ( t ) on the same coordinate axes as I 1 ( t ) .
Solution Summary: The author calculates how much current I 1 is flowing after 0.3 second, 7.5855, and 10.3760, respectively, when time approaches infinity.
119. Current in an RL Circuit The equation governing the amount of current I (in amperes) after time t (in seconds) in a single RL circuit consisting of a resistance R (in ohms), an inductance L (in henrys), and an electromotive force E (in voles) is
(a) If
volts,
ohms, and
henrys, how much current I1 is flowing after 0.3 second? After 0.5 second? After 1 second?
(b) What is the maximum current?
(c) Graph this function
(t), measuring I along the
and t along the
.
(d) If
volts,
ohms, and
henrys, how much current I2 is flowing after 0.3 second? After 0.5 second? After 1 second?
(e) What is the maximum current?
(f) Graph the function
on the same coordinate axes as
.
Instructions.
"I have written solutions in text form, but I need experts to rewrite them in handwriting from A to Z, exactly as I have written, without any changes."
Both in images okk. Instructions.
"I have written solutions in text form, but I need experts to rewrite them in handwriting from A to Z, exactly as I have written, without any changes."
Question 1:
If a barometer were built using oil (p = 0.92 g/cm³) instead of mercury (p =
13.6 g/cm³), would the column of oil be higher than, lower than, or the same as the
column of mercury at 1.00 atm? If the level is different, by what factor? Explain. (5 pts)
Solution:
A barometer works based on the principle that the pressure exerted by the liquid column
balances atmospheric pressure. The pressure is given by:
P = pgh
Since the atmospheric pressure remains constant (P = 1.00 atm), the height of the
liquid column is inversely proportional to its density:
Step 1: Given Data
PHg
hol=hgx
Poil
• Density of mercury: PHg = 13.6 g/cm³
Density of oil: Poil = 0.92 g/cm³
• Standard height of mercury at 1.00 atm: hμg
Step 2: Compute Height of Oil
= 760 mm = 0.760 m
13.6
hoil
= 0.760 x
0.92
hoil
= 0.760 × 14.78
hoil
= 11.23 m
Step 3: Compare Heights
Since oil is less dense than mercury, the column of oil must be much taller than that of
mercury. The factor by which it is taller is:
Final…
Chapter 5 Solutions
Precalculus Enhanced with Graphing Utilities Plus MyLab Math with Pearson eText - Access Card Package (7th Edition) (Sullivan & Sullivan Precalculus Titles)
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