Yes, after a short time that the circuit was connected the capacitor is charged and does not take current any more.
The capacitor is connected parallel to the 4 ohm resistor. What do you know about the voltage across parallel connected circuit elements?
Okay so I can calculate current I=V/R = 19/16 = 1.1875 A
V total = (V of Capacitor) + (V of 11Ω Resistor)
V of 11Ω Resistor = IR = (1.1875)(11) = 13.0625
19 = (V of Capacitor) + 13.0625
V of Capacitor = 5.9375
Then plugging into the capacitance formula:
Q = CV = (4µ)(5.9375) = 23.75µ
Does this work look good?
The current is correct.
No, what do you mean of Vtotal? And you do not need it.
You know the voltage across the 11 Ω resistor. Look at the figure. Where are the plates of the capacitor connected to?
Is the voltage I want then just the voltage across the 4Ω resistor?
Yes. And what current flows though it?
Ah is it the same current as throughout, since the capacitor is quickly charged?
(I = 1.1875 A so V = (1.1875)(4) and then this voltage is the same that would go through the capacitor?)
Yes, they are connected to the same points, so the voltage must be the same across both. (The voltage does not "go through", it is across two points)
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