What is the capacitance of a capacitor if 10 muC of charge are present when 100
ID: 1832720 • Letter: W
Question
What is the capacitance of a capacitor if 10 muC of charge are present when 100 V are applied 1) accross its plates? 1000 muF 0.1 muF 107 muF 10 muF See Figure 10.1 What is the time constant t for this circuit? 50 ms 500 ms 5 ms 5 s See Figure 10.1 if the initial voltage of the capacitor = 0V, after the closung of the switch, when will the voltage VC reach 5.0 V? 1.5 s 0.5 s 0.35 s It will never reach 5.0 V. See Figure 10.1 if the initial voltage of the capacitor = 0V, what is the amxium instaneous current that will flow through the capacitor after the closing the swith? 1 mA 200 kA 5 muA infinityExplanation / Answer
1.) This is simply C=Q/V.
C= 10uC / 100V = .1uF
Ans: B
2.) When a capacitor is connected in a circuit with a resistor, the amount of time needed to charge the capacitor (i.e. the RC time constant) is simply the product of R and C. This value (t) is ONE TIME CONSTANT. Five (5) time constants are requried to charge the capacitor.
t= R x C
t= 10,000 x 50uF = 500ms
Ans: B
3.) During each "time constant", the voltage across the capacitor grows to 63.2% of the maximum amount that it can change. in this circuit, the capacitor will charge until it has "stored" 10V. At which time, the voltage across the resistor will be 10v-10v=0v. no voltage means no current.
So, if it takes .5 seconds (or 500ms) for the capacitor to charge to 63.2% of the value it can reach, then Vc = 6.32 Volts @ t = 500ms.
But the question asks when Vc = 5 volts, not 6.32 volts
(.5 seconds / 6.32 volts) = (X seconds / 5 volts)
X = .5 seconds * 5 volts / 6.32 volts = .395 seconds.
I can't quite read the available answers, but I believe Ans: C
4.) At the initial closing of the switch, the capacitor has no field stored yet. So the only elements in teh circuit are the voltage source and the resistor.
10 V / 10kOhms = 1 mA
Ans: A
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