Respuesta :
Answer:
a) 0 b) 272 μC . c) 446 μC d) 630 μC d) 756 μC
Explanation:
a) at t=0, as the voltage through a capacitor can't change instantanously, Vc must be 0.
By definition, the charge on a plate of a capacitor, the capacitance, and the voltage between the plates, are related by this expression:
C= Q/V, so if V=0, Q must be 0 also, due to C is a constant only dependent upon the geometry and the dielectric material between plates.
b) As current starts to flow, the charge begins to build upon the plates, generating a voltage that opposes to the applied voltage, decreasing the value of the current.
The charge on any plate of the capacitor, at any time, can be written as follows:
Q = CV (1- e⁻t/RC) so, at t= 5s, ⇒ Q= 272 μC
c) At t=10s, applying the same expression, we have:
Q= 446 μC
d) At t= 20s, using the same formula:
Q= 630 μC
e) When t reaches to 100 s, the capacitor is fully charged, reaching to the maximum possible value, CV, which is equal to 756 μC .
In this condition, no current remains flowing, as the capacitor has developed a voltage of the same value (and opposite sign) to the applied voltage.
The charge on the capacitor in the equation at the various given times are;
- at t = 0; Q = 0
- at t = 5 s; Q = 271.96 μC
- at t = 10 s; Q = 446.08 μC
- at t = 20 s; Q = 628.95 μC
- at t = 100 s; Q = 756 μC
A) The formula for the charge on the plate of a capacitor is given by;
Q = CV
Now at a time of 0 seconds connection, it means voltage can't change yet and as a result V = 0 and as such Q = 0 too.
B) The formula for the charge on the plate of the capacitor, at any given time(t) after connection is given by the relationship;
Q = CV(1 - e^(⁻t/RC))
We are given;
C = 12.6 μF = 12.6 × 10⁻⁶ F
R = 0.890 MΩ = 0.89 × 10⁶ Ω
V = 60 V
Thus; at t= 5s,
Q = 12.6 × 10⁻⁶ × 60(1 - (e^(⁻5/(12.6 × 10⁻⁶ × 0.89 × 10⁶))
⇒ Q = 271.96 × 10⁻⁶ C
⇒ 271.96 μC
c) At t=10s;
Q = 12.6 × 10⁻⁶ × 60(1 - (e^(⁻10/(12.6 × 10⁻⁶ × 0.89 × 10⁶))
Q = 446.08 × 10⁻⁶ C
Q = 446.08 μC
d) At t = 20s;
Q = 12.6 × 10⁻⁶ × 60(1 - (e^(⁻20/(12.6 × 10⁻⁶ × 0.89 × 10⁶))
Q = 628.95 × 10⁻⁶ C
Q = 628.95 μC
e) At t = 100 s;
Q = 12.6 × 10⁻⁶ × 60(1 - (e^(⁻100/(12.6 × 10⁻⁶ × 0.89 × 10⁶))
Q = 756 μC
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