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.

(a) 0.075 s

The time constant of a RC circuit is given by:

where

R is the resistance of the circuit

C is the capacitance

In this problem, we have

So the time constant is

(b)

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

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

(c)

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

where

is the capacitance

is the voltage drop across the capacitor

Substituting,

(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:

where

is the maximum value of the current

is the time constant

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

Solving for t, we find

(f)

The charge on the capacitor is given by

where

Substituting into the equation, we find

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

answer(20) psi;