PLEASE PUT IN PROPER UNITS!! 1) The switch has been open for a long time when at

Question

PLEASE PUT IN PROPER UNITS!!

1) The switch has been open for a long time when at time t = 0, the switch is closed. What is I4(0), the magnitude of the current through the resistor R4 just after the switch is

closed?

.06995 A

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2) What is Q(?), the charge on the capacitor after the switch has been closed for a very long time?

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3) After the switch has been closed for a very long time, it is then opened. What is Q(topen), the charge on the capacitor at a time topen = 518 ?s after the switch was opened?

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4) What is IC,max(closed), the current that flows through the capacitor whose magnitude is maximum during the time when the switch is closed? A positive value for the current is defined to be in the direction of the arrow shown.

A

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5) What is IC,max(open), the current that flows through the capacitor whose magnitude is maximum during the time when the switch is open? A positive value for the current is defined to be in the direction of the arrow shown.

A

Explanation / Answer

Knowns:

V = 12V
R? = 33?
R? = 33?
R? = 71?
R? = 116?
C = 65?F

(1) You should consider the capacitor to be in the "discharged" state just prior to t=0. Therefore, the voltage across it is 0V. This causes R? to be effectively in parallel with R?. That parallel combination is in series then with R? and R?. Given that, you can compute the current.

Itotal = I? = I? = V ? (R?+R?+(R? || R?))
I? = 12V ? (33?+116?+(33? || 71?))
I? = 12V ? (33?+116?+22.52?)
I? = 12V ? 171.52? ? 69.95mA

(2) The capacitor, at t=?, has no more currents leaving or arriving as everything has reached equilibrium by then. R?'s current is also therefore zero and so R? doesn't drop any voltage. Without current through R? and C, the capacitor can be simply removed without affecting the circuit (or treated as an ? resistance.) And so can R?. This leaves only R?, R?, and R? to worry about. First get the currents:

Itotal = I? = I? = I? = V ? (R?+R?+R?)
I? = 12V ? (33?+71?+116?)
I? = 12V ? 220? ? 54.54mA

The voltage drop across R? will be the voltage across C:

V? = I? ? R? = 54.54mA ? 71? ? 3.872V

(3) The only components to worry about with the switch open are C, R? and R?. The starting voltage on C is V? = 6.361V, from above. You should know that:

V(t) = V? ? e^(-t ? ((R?+R?)?C))
V(t=528?s) = 6.361V ? e^(-528?s ? ((25?+97?)?42?F))
V(528?s) = 6.361V ? e^(-528?s ? ((25?+97?)?42?F))
V(528?s) ? 5.738V

I'm actually going to stop here. I've done enough damage already.

EDIT: Okay, since you ask J. I'll do the 4th.

(4) The maximum current into C, Ic(max), will occur when the switch is first closed, at t=0. [Taking the assumption that Vc=0 at t=0, as already mentioned in (1) above.] This takes us back to the situation in (1), in fact. We already calculated the loop current as 67.46mA. But R? "looked" to be in parallel with R?, since Vc=0 at t=0. So this 67.46mA splits between R? and R?.

I(R?@t=0) ? 19.877? ? 25? ? 67.46mA ? 53.64mA
I(R?@t=0) ? 19.877? ? 97? ? 67.46mA ? 13.82mA

It's the case that Ic(max) = I(R?@t=0) ? 53.64mA.

You could compute it another way. Since R? and R? are in series, replace them with a single 158? resistor (short R? and set R?=158?.) Now consider the (-) terminal of V to be the GND (0V) node. Convert V, the new R?, and R? into their The?venin equivalent:

Vth = V?R? ? (R?+R?) = 12V?97? ? (158?+97?) ? 4.5647V
Rth = R??R? ? (R?+R?) ? 60.102?

Rth is now in series with R? and Vc=0V at t=0, so all of V appears across this new series pair. The current is:

Ic(max) = Vth ? (Rth+R?) = 4.5647V ? (60.102?+25?) ? 53.64mA

Same answer, different method.

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