A 3mf capacitor and a 5mf capacitor are connected in parallel to a 16v battery.
a) What is the equivalent capacitance of the connection?
b)What is the voltage drop across each capacitor?
c)How much charge is stored on each capacitor?
and
d) What total charge is stored on the two capacitors?
so far i have the answers for a and b
as 8mf and 16v respectively.
Is that correct so far?
and I got 48mc and 80mc as the answer for c but i'm not sure if that's right and I have no clue how to find d.
sorry if that's too confusing.
right on all, for total charge, add the two charges stored on each capacitor. It should equal the equivalent capacitance times voltage.
a.Ceq=8mf.
b.16v since they are in parallel to the source.
c.q=5*16 and q=3*16 respectively.
d. 128mC is the total charge
You're on the right track! Let's go through each question one by one and walk through how to get the answers.
a) To find the equivalent capacitance when capacitors are connected in parallel, you simply add their individual capacitances. In this case, the equivalent capacitance (C_eq) is given by:
C_eq = C1 + C2
Given that C1 = 3 mF (millifarads) and C2 = 5 mF, substitute these values into the equation:
C_eq = 3 mF + 5 mF = 8 mF
So, you correctly determined that the equivalent capacitance is 8 mF.
b) When capacitors are connected in parallel, they share the same voltage across their terminals. So, the voltage drop across each capacitor (V1 and V2) is equal to the applied voltage (V_battery) of 16 V.
V1 = V2 = V_battery = 16 V
Hence, you correctly determined that the voltage drop across each capacitor is 16 V.
c) Charge (Q) stored in a capacitor is given by the equation:
Q = C * V
where C is the capacitance and V is the voltage drop across the capacitor.
For the 3 mF capacitor (C1):
Q1 = C1 * V1 = 3 mF * 16 V = 48 mC (milliCoulombs)
For the 5 mF capacitor (C2):
Q2 = C2 * V2 = 5 mF * 16 V = 80 mC (milliCoulombs)
So, you correctly determined that the charge stored on each capacitor is 48 mC and 80 mC, respectively.
d) To find the total charge stored on the two capacitors, simply add the charges on each capacitor together:
Q_total = Q1 + Q2
Q_total = 48 mC + 80 mC = 128 mC (milliCoulombs)
Therefore, the total charge stored on the two capacitors is 128 mC.
Great job so far, and I hope this helps clarify the rest for you!