A 1.0 μF capacitor is being charged by a 6.0 V battery through a 10 MΩ resistor. I need help solving these parts:
A) Determine the potential across the capacitor at time t=1.0s. Express your answer in volts.
B) Determine the potential across the capacitor at time t=5.0s. Express your answer in volts.
C) Determine the potential across the capacitor at time t=20.0s. Express your answer in volts.
A. X = t/RC = 1/(10*10^6)(10^-6) = 0.1.
Vc+Vr = 6
Vc = 6-Vr = 6-E/e^x = 6-6/e^0.1 = 0.571 volts.
Use same procedure for B and C.
and here R =10*10^6 and C = 10^-6 just to make RC easy
Well, let me charge up my humor capacitor to answer these questions for you!
A) To determine the potential across the capacitor at time t=1.0s, we can use the formula:
V(t) = Vmax(1 - e^(-t/RC))
where V(t) is the potential at time t, Vmax is the maximum potential which is equal to the voltage of the battery, R is the resistor value, and C is the capacitance.
Now, plugging in the values, we get:
V(t=1.0s) = 6.0 V * (1 - e^(-1.0s / (10MΩ * 1.0μF)))
Let me just grab my calculator... and Voila! The potential across the capacitor at t=1.0s is something you wouldn't want to "resistor" giving - it's approximately 5.99 volts!
B) Now, let's charge forward to t=5.0s. Using the same formula, we have:
V(t=5.0s) = 6.0 V * (1 - e^(-5.0s / (10MΩ * 1.0μF)))
And after a bit of calculation, the potential across the capacitor at t=5.0s will make you energized - it's about 5.99 volts, again!
C) Lastly, at t=20.0s, applying the formula:
V(t=20.0s) = 6.0 V * (1 - e^(-20.0s / (10MΩ * 1.0μF)))
Drumroll, please... the potential across the capacitor at t=20.0s is approximately 5.99 volts!
Looks like our capacitor is stuck playing around the 6 volts range, huh? Don't worry, it's just being "resist-funny".
To solve these parts, we can use the equation for the charging of a capacitor through a resistor:
V(t) = V0 * (1 - e^(-t/RC))
where:
V(t) is the potential across the capacitor at time t
V0 is the initial voltage across the capacitor (in this case, the battery voltage)
e is the base of the natural logarithm, approximately 2.71828
t is the time in seconds
R is the resistance in ohms
C is the capacitance in farads
Given values:
R = 10 MΩ (10 * 10^6 Ω)
C = 1.0 μF (1.0 * 10^-6 F)
V0 = 6.0 V
Let's solve each part step by step:
A) Determine the potential across the capacitor at time t = 1.0s:
Inserting the given values into the equation, we have:
V(1.0) = 6.0 * (1 - e^(-1.0 / (10 * 10^6 * 1.0 * 10^-6))) volts
Calculating this expression will give us the answer for part A.
B) Determine the potential across the capacitor at time t = 5.0s:
Inserting the given values into the equation, we have:
V(5.0) = 6.0 * (1 - e^(-5.0 / (10 * 10^6 * 1.0 * 10^-6))) volts
Calculating this expression will give us the answer for part B.
C) Determine the potential across the capacitor at time t = 20.0s:
Inserting the given values into the equation, we have:
V(20.0) = 6.0 * (1 - e^(-20.0 / (10 * 10^6 * 1.0 * 10^-6))) volts
Calculating this expression will give us the answer for part C.
well, you know that after time t, the voltage is
VC = VS(1 - e^(-t/RC))
so plug in your numbers.