Under what conditions can an ideal gas undergo a change of state without doing external work?
1 answer:
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In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
where t is the time,
is the maximum charge on the capacitor at voltage V, and 
is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.
1) Using
and 
, the time constant of the circuit is:
2) To find the charge on the capacitor at time
, we must find before the maximum charge on the capacitor, which is
And then, the charge at time is equal to
3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be
. We can see this also from the previous formule, by using 
:
where t is the time,
Using this law, we can answer all the three questions of the problem.
1) Using
2) To find the charge on the capacitor at time
And then, the charge at time
3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be
Answer:
x(t) = -8sin2t
Explanation:
See the attachment for solution
From my solving, we can deduce that w² = 4, and thus, w = 2
Therefore, the general solution is
x(t) = c1 cos2t + c2 sin2t
Considering the final variable, we can conclude that
x(0) = 0
x'(0) = -8 m/s
The final solution, thus
x(t) = -8sin2t
Answer:
The final temperature of the element = 262.67°C
The power dissipated in the heating element initially = 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W
Explanation:
Our given parameters include;
A Nichrome heating element operates on 120 V.
Voltage (V) = 120V
Initial Current (I₁) = 1.36 A
Initial Temperature (T₁) = 28°C
Final Current (I₂) = 1.23 A
Final Temperature (T₂) = unknown ????
Temperature dependencies of resistance is given by:
---------------------- (1)
in which R₁ is the resistance at temperature T₁
is the resistance at temperature T₂
Given that V= IR
R =
Therefore, the resistance at temperature 28°C is;
= 88.24Ω
=
= 97.56Ω
From (1) above;
97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]
1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)
0.1056 = 4.5×10⁻⁴(T₂-28°C)
T - 28° C = 234.67
T = 234.67 + 28° C
T = 262.67 ° C
(b)
What is the power dissipated in the heating element initially and when the current reaches 1.23 A
The power dissipated in the heating element initially can be calculated as:
P = I²₁R₂₈
P = (1.36A)²(88.24Ω)
P = 163.209 W
P ≅ 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:
P = (1.23)²(97.56Ω)
P = 147.598524
P ≅ 147.60 W