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I want to differentiate the following?
The formula for instantaneous voltage during a capacitor charge is v = V(1 -e^ - (0.006/0.0047)) but I think I am getting the wrong answer. Not homework, genuine problem. Any clues, please, including how you did it?
The answer is the instantaneous voltage after 6ms - the question asks me to differentiate. The value of big V is 5, btw. 0.0047 is the CR time constant
2 Answers
- Anonymous10 years agoFavorite Answer
I assume you are differentiating in respect of V?
Note before we start that (1 -e^ - (0.006/0.0047)) is just a number, so when v = V(1 -e^ - (0.006/0.0047))
dv/dV=(1 -e^ - (0.006/0.0047))≈0.721
May I ask why differentiating helps?
- No MythologyLv 710 years ago
You seem to be missing something here, namely the time variable. If the voltage across the capacitor (DC circuit) is
v(t) = V(1 - e^(-αt))
where α is constant in units 1/sec, and V is a constant in volts, then the rate of change of this is
v'(t) = αVe^(-αt).
If α = 1/0.0047 and t = 0.006, you get
v'(0.006) = 0.0047V e^(-0.006/0.0047).
