# Isaac physics

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#1
Can someone help me with the following questions:
https://isaacphysics.org/questions/t...9-f3de346cfffc
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#2
For part A) I was thinking that for the greatest charging time you need the greatest resistance ( i.e. switches S1 and S2)
For parts B and C i tried using Kirchhoff's laws with the voltage split according to the capacitances.

Ratio c1:c2 is 1:10 hence (since greater capacitance provides lower voltage) voltage across c1 is 2.1818V and c2 is 0.21818V. Subsequently, applying the current and voltage laws.

I1=I2+I3
2.1818=330I1+15I2
0.21818=15I2+1000I3

Could please tell me where I'm going wrong with this. Thanks!
0
3 weeks ago
#3
For part A) I was thinking that for the greatest charging time you need the greatest resistance ( i.e. switches S1 and S2)
For parts B and C i tried using Kirchhoff's laws with the voltage split according to the capacitances.

Ratio c1:c2 is 1:10 hence (since greater capacitance provides lower voltage) voltage across c1 is 2.1818V and c2 is 0.21818V. Subsequently, applying the current and voltage laws.

I1=I2+I3
2.1818=330I1+15I2
0.21818=15I2+1000I3
This question is poorly worded, as, for part A, only when S1 and S2 are closed do both capacitors discharge. What they appear to be after is the longest discharge time for the capacitor(s) that actually discharge.

A: The capacitors are fully-charged, and will be discharging. The way to compare these is to consider each option, and calculate the time constant (RC) of the effective circuit. That with the largest time constant will take the longest to discharge.

B: Calculate how the 2.4V is split between C1 and C2 (Q=CV, V=Q/C, Q is common). You then replace the capacitors with voltage sources, to obtain a simple circuit.

C: Hint: This will occur when the voltage at the mid-point of R1 and R2 is the same as that between C1 and C2. You could also generate an expression for the current through R3 for part B, but you'd need to leave it in terms of C1, which makes it a lot more complicated.
Last edited by RogerOxon; 3 weeks ago
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3 weeks ago
#4
(Original post by RogerOxon)
B: Calculate how the 2.4V is split between C1 and C2 (Q=CV, V=Q/C, Q is common). You then replace the capacitors with voltage sources, to obtain a simple circuit.
Hint: Start with and . You can write as (be careful to label the positive direction for each current). Write equations for the voltage across each capacitor (which you know) in terms of these currents, and solve for .
1
3 weeks ago
#5
I1=I2+I3
2.1818=330I1+15I2
0.21818=15I2+1000I3

Could please tell me where I'm going wrong with this. Thanks!
Signs. Be careful to mark the positive direction for each current - here you have I3 and I2 in different directions around the C2 "loop".
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#6
(Original post by RogerOxon)
Signs. Be careful to mark the positive direction for each current - here you have I3 and I2 in different directions around the C2 "loop".
Thanks for responding! I tried doing it again but with no success...
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3 weeks ago
#7
Thanks for responding! I tried doing it again but with no success...
What Kirchhoff says is that the nett current at a junction is zero. Another way of stating this is that the current in must equal the current out. With the diagram above (after swapping I2 and I3, so that they're through R2 and R3 respectively), your equation should therefore be: In terms of voltages, the drop across C1 is going to be (with the diagram above, and the I2/I3 swap): Last edited by RogerOxon; 3 weeks ago
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#8
(Original post by RogerOxon)
What Kirchhoff says is that the nett current at a junction is zero. Another way of stating this is that the current in must equal the current out. With the diagram above (after swapping I2 and I3, so that they're through R2 and R3 respectively), your equation should therefore be: In terms of voltages, the drop across C1 is going to be (with the diagram above, and the I2/I3 swap): Got it, cheers
0
3 weeks ago
#9
Got it, cheers
Good. Signs can be annoying 1
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