Respuesta :
(a) 0.075 s
The time constant of a RC circuit is given by:
[tex]\tau=RC[/tex]
where
R is the resistance of the circuit
C is the capacitance
In this problem, we have
[tex]R=11.5 k\Omega=1.15\cdot 10^4 \OmegaC=6.5\mu F=6.5\cdot 10^{-6}F[/tex]
So the time constant is
[tex]\tau=RC=(1.15\cdot 10^4 \Omega)(6.5\cdot 10^{-6} F)=0.075 s[/tex]
(b) [tex]V_C = 25 V[/tex]
Initially, the capacitor is uncharged. When the switch is closed:
- at the very first instant, the voltage drop across the resistor ([tex]V_R[/tex]) is equal to the voltage of the battery ([tex]V_B=25 V[/tex]), while the voltage drop across the capacitor is zero
- Then current starts to flow in the circuit, and so the charge on the capacitor increases: as a consequence, the voltage drop across the capacitor, [tex]V_C[/tex], increases as well, while the voltage drop across the resistor decreases
- After a very long time, the capacitor has been completely charged, so that the voltage drop across the capacitor has become equal to the voltage of the battery: [tex]V_C = V_B =25 V[/tex], while the potential drop across the resistor is zero and no more current flows in the circuit.
(c) [tex]1.63\cdot 10^{-4} C[/tex]
As we said at point b), a very long time after the switch has been closed, the capacitor is fully charged, and so its charge is given by
[tex]Q_0=CV[/tex]
where
[tex]C=6.5\cdot 10^{-6}F[/tex] is the capacitance
[tex]V=25 V[/tex] is the voltage drop across the capacitor
Substituting,
[tex]Q_0=(6.5\cdot 10^{-6} F)(25 V)=1.63\cdot 10^{-4} C[/tex]
(d) Zero
As we said at point b), a very long time after the switch has been closed, no more current is flowing through the circuit, because the capacitor is now fully charged (so the potential drop across it is equal to the voltage of the battery) and cannot store more charge; on the contrary, the voltage drop across the resistor is now zero, so no current can flow through the circuit.
(e) 0.082 s
The current through the resistor is given by:
[tex]I(t) = I_0 e^{-\frac{t}{\tau}}[/tex]
where
[tex]I_0 = \frac{V_b}{R}[/tex] is the maximum value of the current
[tex]\tau=0.075 s[/tex] is the time constant
We want to find the time t at which the current is 1/3 of the maximum value [tex]I_0[/tex], which means:
[tex]\frac{I(t)}{I_0}=e^{-\frac{t}{\tau}}=\frac{1}{3}[/tex]
Solving for t, we find
[tex]t=-\tau ln(\frac{1}{3})=-(0.075 s)(ln\frac{1}{3})=0.082 s[/tex]
(f) [tex]1.1\cdot 10^{-4}C[/tex]
The charge on the capacitor is given by
[tex]Q(t)=Q_0 (1-e^{-\frac{t}{\tau}})[/tex]
where
[tex]Q_0 = 1.63\cdot 10^{-4} C\\t = 0.082 s\\\tau = 0.075 s[/tex]
Substituting into the equation, we find
[tex]Q(t)=(1.63\cdot 10^{-4}C) (1-e^{-\frac{0.082 s}{0.075 s}})=1.1\cdot 10^{-4}C[/tex]
