'''Discussion Question 7A''' '''P212, Week 7''' ''RC Circuits'' The circuit shown initially has the capacitor uncharged, and the switch connected to neither terminal. At time ''t'' = 0, the switch is thrown to position a. {| | |
''C''
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a
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b
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'''E''' = 12 V
| |- | '''E''' | |
''R''2
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''C'' = 5 µF
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''R''1
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''R''1
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= 3 Ω
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''R''2
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= 6 Ω
| |} [[File:media/image1.jpeg|334x151px]]

  1. '''At ''t'' = 0+, immediately after the switch is thrown to position a, what are the currents ''I''1 and ''I''2 across the two resistors?'''

What does the uncharged capacitor ''look like'' to the rest of the circuit at time 0? Does it offer ''any'' resistance to the flow of charge? (Why or why not?) At t = 0, the capacitor acts like wire w/ no resistance to current flow. Hence the battery is effectively hooked to ''R''2 = 6Ω ⇒ ''I'' = E/ ''R''2 = 12''V'' / 6Ω = 2 A
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  1. '''After a very long time, what is the instantaneous power ''P'' dissipated in the circuit?''' After a very long time, what will have happened to the capacitor? ''Now'' what will it look like to the rest of the circuit?

After a long time the capacitor acts like an open circuit and the battery current flows through ''R''1 + ''R''2 . Hence I = E /(''R''1 + ''R''2 .) = 12/9 =4/3 A and the power is
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  1. '''After a very long time, what is the ''Q'' charge on the capacitor?''' To detemine ''Q'', you need the voltage across the capacitor ...

From part (b), we know after a long time I=4/3 A. The voltage drop across the capacitor is the same as the voltage drop across R1 → ∆''V''cap =''IR''1 =4/3 A × 3Ω = 4''V'' .
[[File:media/image4.jpeg|570x76px]] Next, after a very long time ''T'', the switch is thrown to position b.
''C''
[[File:media/image5.jpeg|334x151px]]
a
{| | '''E''' |
b
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''R''2
| |- | | | | |- | | |
''R''1
| |} '''(d) What is the time constant τ that describes the discharging of the capacitor?'''
We have a nice formula available for time constants: τ = ''RC''. But the ''R'' in the formula refers to the ''total resistance through which the capacitor discharges''. Redrawing your circuit might help you to determine this ''R'' . When the switch is thrown to b, the capacitor discharges through R1 and R 2 in parallel. Requiv = R1 R2 / ( R1 + R2 ) = 3(6) /(3 + 6) = 2Ω ; ''τ'' = ''CR''equiv = ( 5''µ'' ''F'' )( 2Ω ) = 10''µs''
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  1. '''Write down an equation for the time dependence of the charge on the capacitor, for times ''t'' > ''T''. Your answer for ''Q''(''t'') should depend only on the known quantities E, ''R''1, ''R''2, ''C'','''

'''and ''T''.'''
You know the general form for the time dependence of a discharging capacitor. All you have to do is fix the constants in this expression to match the charge at ''t'' = ''T'' and at ''t'' = ∞. We know that Q will involve a exp(-t/''τ'' ) factor added to a possible constant. The boundary conditions are at ''t'' = ''T'' + ,''Q'' = 20 ''µC'' and fades to zero at infinity.
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The form that does this is ''Q'' = 20''µC'' exp − ''t'' − ''T'' 10''µs''
'''(f) What is the charge ''Q''20 on the capacitor 20 µsec after time ''T''?''' {| |
''Q'' = 20''µC'' exp −
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''t'' − ''T''
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= 20''µC'' exp −
| 20''µs'' |
= 20e−2 = 2.71 ''µ''C
| |- | | | | 10''µs'' | | |- |
10''µs''
| | | | |} '''(g) What is the current through R2 20 µsec after time ''T'' ?'''
The voltage across R2 is the ∆''Vcap'' = ''Q''20 / ''C'' = 2.71''µ'' ''C'' / 5''µF'' = 0.542 ''V'' ∆''Vcap'' = ''I'' 2 ''R''2 → ''I'' 2 = ( 0.542 ''V'' ) / ( 6Ω ) = 0.0903 A
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