So this exam question looked straight forward, but I can't get the right answer.

I'm trying to find the time, from when the switch is close,that the current will equal 2mA. I've used a few formulas but none give the right answer.

What am I doing wrong?

Also the tau and time confuse me a bit.

Any help would be greatly appreciated.

My calculations

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    \$\begingroup\$ You need to include a schematic. Without it we don't know how the switch is connected. \$\endgroup\$ – The Photon Jun 10 '15 at 17:37
  • \$\begingroup\$ Sorry I forgot I fixed it now. \$\endgroup\$ – Omuse Jun 10 '15 at 17:49

This is a rather poorly set question- no information is given as to the initial condition of the capacitor. If the 10uF cap happens to be charged to 12V the current will be 0 before and after the switch closes.

Anyway, what they want you to do is to assume the initial voltage on the capacitor is 0V.

You should then be able to write down the equation for voltage on the capacitor as a function of time (memorize this- it's the solution to the differential equation if you want to do it from first principles). $$ V_c(t) = 12 (1 - e^{-t/\tau}) $$ where \$\tau = RC\$.

Since you know that the current is \$(12V - V_c(t)) \cdot 2.2k\$ you can solve for when current is 2mA.

Alternately you can recognize that $$i(t) = I_0 e^{-t/\tau}$$ where \$I_0 = 12V/2.2K\$ (you have this)

so $$ t = -\text{ln}(2mA/I_0)* RC $$ and the answer is 'b' 22ms

To get the above line, divide both sides by \$I_0\$, then take the ln() of both sides and solve for t.

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    \$\begingroup\$ Thank you for the help I got the right answer this time. \$\endgroup\$ – Omuse Jun 11 '15 at 9:29

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