Physics, 21.01.2020 08:31 granthazenp5e9mj

Acapacitor of capacitance c = 6.5 μf is initially uncharged. it is connected in series with a switch of negligible resistance, a resistor of resistance r = 11.5 kω, and a battery which provides a potential difference of vb = 25 v. show answercorrect answer17% part (a) calculate the time constant τ for the circuit in seconds. show answer correct answer17% part (b) after a very long time after the switch has been closed, what is the voltage drop vc across the capacitor in terms of vb? show answer correct answer17% part (c) calculate the charge q on the capacitor a very long time after the switch has been closed in c. no attempt no attempt17% part (d) calculate the current i a very long time after the switch has been closed in a. no attempt no attempt17% part (e) calculate the time t after which the current through the resistor is one-third of its maximum value in s. show answerno attempt17% part (f) calculate the charge q on the capacitor when the current in the resistor equals one third its maximum value in c.

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Answer from: chem1014

(a) 0.075 s

The time constant of a RC circuit is given by:

$\tau=RC$

where

R is the resistance of the circuit

C is the capacitance

In this problem, we have

$R=11.5 k\Omega=1.15\cdot 10^4 \OmegaC=6.5\mu F=6.5\cdot 10^{-6}F$

So the time constant is

$\tau=RC=(1.15\cdot 10^4 \Omega)(6.5\cdot 10^{-6} F)=0.075 s$

(b) V_C = 25 V

Initially, the capacitor is uncharged. When the switch is closed:

- at the very first instant, the voltage drop across the resistor ( V_R ) is equal to the voltage of the battery ( V_B=25 V ), 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, V_C , 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: V_C = V_B =25 V , while the potential drop across the resistor is zero and no more current flows in the circuit.

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

Q_0=CV

where

$C=6.5\cdot 10^{-6}F$ is the capacitance

V=25 V is the voltage drop across the capacitor

Substituting,

$Q_0=(6.5\cdot 10^{-6} F)(25 V)=1.63\cdot 10^{-4} C$

(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:

$I(t) = I_0 e^{-\frac{t}{\tau}}$

where

$I_0 = \frac{V_b}{R}$ is the maximum value of the current

$\tau=0.075 s$ is the time constant

We want to find the time t at which the current is 1/3 of the maximum value I_0 , which means:

$\frac{I(t)}{I_0}=e^{-\frac{t}{\tau}}=\frac{1}{3}$

Solving for t, we find

$t=-\tau ln(\frac{1}{3})=-(0.075 s)(ln\frac{1}{3})=0.082 s$

(f) $1.1\cdot 10^{-4}C$

The charge on the capacitor is given by

$Q(t)=Q_0 (1-e^{-\frac{t}{\tau}})$

where

$Q_0 = 1.63\cdot 10^{-4} C\\t = 0.082 s\\\tau = 0.075 s$

Substituting into the equation, we find

$Q(t)=(1.63\cdot 10^{-4}C) (1-e^{-\frac{0.082 s}{0.075 s}})=1.1\cdot 10^{-4}C$

Answer from: Quest

The low lands i believe

Answer from: Quest

when heated, the valance electrons increased their speed and transferred the heat from electron to electron. i just took the test.

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Acapacitor of capacitance c = 6.5 μf is initially uncharged. it is connected in series with a switch...

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