Consider three capacitors in parallel, each with a value of 25.0uF. Also consider an Ammeter placed on the first wire coming out of the positive power feed, breaking the circuit there (this then goes back into the circuit, to the positive side of the capacitors). A potential difference of 4200 V is then established through the circuit. How many coulombs pass through the ammeter? Can you explain your answers?
2007-03-19 17:53:32 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
You circuit description is not clear. Do you mean that the ammeter is inserted into the positive lead of the first capacitor? If that is so, the ammeter only reads the current flow into the first capacitor, and the total charge on that capacitor will be 2500v * 25*10^-6F. All that had to flow through the ammeter.
2007-03-19 18:29:23 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋