Recent content by Teptip

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?)

Is the voltage I want then just the voltage across the 4Ω resistor?

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) =...

Thanks, solved!