chargecurrenttime
This should be pretty basic but I want to clarify some:
I = dQ/dt, so that means that I is the derivative of q(t) right?
So if I'm given q(t) = (7t^2 – 2t +3) pC and told that time is dimensionless for simplicity.
So to solve it I would take the derivative and get 14t-2 (units? pC/time?) Am I doing this right? How would I determine the unit for the answer if t is dimensionless?
Some of the things you say are conflicting, so I want to be clear about the question I am answering. If this is not the question you intended, then you need to write a better question.
The question is what is the voltage accross C1 over time when it initially starts at 0. You didn't specify a capacitance, so we'll leave that as the variable C.
There are two separate regimes to the C1 voltage over time. The first is when the supply is in current limit mode. In that case, the capacitor is being charged up linearly. When the capacitor reaches 9 V, the supply switches over to constant voltage operation. After that, there is a exponential decay from 9 V to 10 V governed by the RC time constant.
The voltage rise of a capacitor is dV = I dT / C. We know the current (I) is 1 Amp and that dV will be 9 V in the first part of the function. The time to reach 9 V is dT = dV C / I. For example, if C = 470 µF, then the time to charge from 0 to 9 V is 4.2 ms. The capacitor voltage will rise linearly during that time.
From 9V on, the supply will be at a constant voltage of 10 V. This remaining 1 V rise will occur as a exponential according to the time constant RC. Again using 470 µF as example for C, that time constant would be 470 µs. That means, for example, that 470 µs after the capacitor has reached 9 V, it will have gained another 630 mV, which would put it at 9.63 V in absolute terms.
Work backwards and assume the supply is always putting out 10 V. At what capacitor voltage does that require 1 A or more? Since R1 is 1 Ω, 1 A thru it causes a 1 V drop. If the supply is 10 V and R1 drops 1 V, then there must be 9 V on the capacitor when the current is 1 A. If the capacitor voltage is lower, then the voltage on R1 must be higher, which means the supply has to source more than 1 A. However, we know the supply puts out the lesser of 10 V or 1 A, so delivering more than 1 A is not possible. This means for capacitor voltages below 9 V, the supply voltage will be the capacitor voltage plus 1 V, since 1 V will be accross R1 when the supply is putting out 1 A.
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.
Best Answer
If you are not given units of time then you can't calculate amperes. If the units in your problem are actually seconds then the units for your answer would be pA. Without a unit for time I guess the best you could say is pC/time, as you did. I don't understand how saying that time is dimensionless makes any sense or simplifies the problem.