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enter image description here

C1 = 56uF C2 = 12uF R = 12kohm E = 13 V

I have this circuit and I want help to find out how long in takes for the voltage Uc2 to achieve the value 7 V?

I have calculated the time constant to 117 ms, and I tried to find the time through t = 0.117 * ln(13/13-7) but that doesn't seem to be right. Could someone help me to go to the right direction?

Thank you!

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1 Answer 1

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In your circuit:

enter image description here

Since the current in a series circuit is everywhere the same at any instant, when S1 closes, the accumulation of charge in C1, in time, will be equal to the accumulation of charge in C2, but since their capacitances are different the voltages across them will be different. .

Then, since \$ Q=CV\$, and E3 is given as 7 volts,

for C2, $$ Q = C2 \times E3 =12\mu F \times 7V =84\mu C $$

and for C1, $$ V=\frac {Q}{C} = \frac{84\mu C}{56\mu F} = 1.5\text{ volts.}$$

The voltage across C2 is given as 7 volts, so E2 will be equal to 7 volts plus the voltage across C1, 1.5 volts, so $$ E2 = E3+1.5V = 8.5V $$

C1 and C2 are in series, so their equivalent capacitance is:

$$C_T = \frac{C1 \times C2}{C1+C2} \approx 9.9 \text{ microfarads.}$$

Our circuit now looks like this:

enter image description here and we can get the time taken to charge Ct to 8.5 volts (which is the time taken to charge C2 to 7 volts) by solving:

$$\tau = k \ RC, $$

where

$$ k = ln\frac{V}{V-V_T} = ln\frac{13V}{13V-8.5V} =1.06$$

which works out to:

$$\tau = 1.06\times\ 1.2\cdot10^{4}\Omega\times 9.9\cdot10^{-6}F \approx 126\text{ milliseconds}. $$

The proof:

enter image description here

and the LTspice circuit list:

Version 4
SHEET 1 880 680
WIRE 144 48 128 48
WIRE 224 48 144 48
WIRE 400 48 304 48
WIRE 480 48 400 48
WIRE 400 80 400 48
WIRE 400 176 400 144
WIRE 480 176 400 176
WIRE 128 208 128 48
WIRE 400 208 400 176
WIRE 128 304 128 288
WIRE 400 304 400 272
WIRE 400 304 128 304
WIRE 128 352 128 304
FLAG 128 352 0
FLAG 480 176 E3
FLAG 480 48 E2
FLAG 144 48 E1
SYMBOL voltage 128 192 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value PULSE(0 13 0 100n)
SYMBOL cap 384 80 R0
SYMATTR InstName C1
SYMATTR Value 56µ
SYMBOL cap 384 208 R0
SYMATTR InstName C2
SYMATTR Value 12µ
SYMBOL res 320 32 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 12k
TEXT 136 328 Left 2 !.tran 150m startup uic
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  • \$\begingroup\$ Hi! Thank you for answering! Do you mean that the time constant that I calculated on the previous question is wrong, should it be 92 milliseconds? Or do you mean that that is the time constant that I should use in t = τ * ln(13/13-7)? :) \$\endgroup\$
    – Vetenskap
    Apr 24, 2016 at 17:54
  • \$\begingroup\$ I'm sorry, I forgot to mention that when I calculated the time for Uc2 to reach 7 V, then I calculated: 117*0.773 which gave me about 90 ms, but that is wrong (92ms is also wrong) when I write it in the system I'm using when checking if my answer is right \$\endgroup\$
    – Vetenskap
    Apr 24, 2016 at 18:03
  • \$\begingroup\$ Oops... I misread your question. I thought you were talking about the voltage from the junction of R and C1 to ground. I'll fix it in a while. \$\endgroup\$
    – EM Fields
    Apr 24, 2016 at 18:56
  • \$\begingroup\$ It's okay, it was my fault that I wasn't clear enough. :) \$\endgroup\$
    – Vetenskap
    Apr 24, 2016 at 19:02
  • \$\begingroup\$ My pleasure. If you found my answer helpful, and better answers don't show up in a few days, I'd appreciate your accepting my answer, which will increase my reputation (status here) by 15 points. \$\endgroup\$
    – EM Fields
    Apr 25, 2016 at 23:27

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