A 6.30 μF capacitor that is initially uncharged is connected in series with a 4300 Ω resistor and a 501 V emf source with negligible internal resistance.
I need help on these two questions:
1) A long time after the circuit is completed (after many time constants), what are the voltage drop across the capacitor and across the resistor?
Express your answers in volts separated by a comma.
2) A long time after the circuit is completed (after many time constants), what is the current through the resistor?
Express your answer in amperes.
1. After 5 Time constants, the capacitor is fully charged and the current is zero.
Therefore, Vc = 501 volts. Vr = 0 volts.
2. I = (E-Vc)/R = (501-501)/4300 = 0 Amperes.
1) Ah, the long-awaited time after many time constants! The voltage drop across the capacitor can be simply described as "I'm feeling positively shocked!" - because it reaches the same potential as the emf source, which is 501 volts. As for the voltage drop across the resistor, it can be summed up as "You can always rely on me to resist, and the potential difference will be 501 volts too!" So, the voltage drop across the capacitor is 501 volts, and the voltage drop across the resistor is also 501 volts.
2) After many time constants have passed, the current through the resistor can be measured as "A whopping zappy current of 0.1165 Amperes!"
To answer the questions, we can use the concepts of RC circuits and time constants. The time constant (τ) of an RC circuit is given by the product of the resistance (R) and the capacitance (C), τ = RC.
1) The voltage drop across the capacitor (Vc) and across the resistor (Vr) after many time constants can be determined using the formula:
Vc = emf (1 - e^(-t/τ))
Vr = emf * e^(-t/τ)
Since the circuit is completed for a long time, we can assume that t is much larger than τ, therefore the exponential term e^(-t/τ) approaches zero.
Substituting the given values:
C = 6.30 μF = 6.30 x 10^(-6) F
R = 4300 Ω
emf = 501 V
Calculating the time constant:
τ = RC = (6.30 x 10^(-6) F) * (4300 Ω) = 27.09 seconds
For a long time after the circuit is completed:
Vc ≈ emf = 501 V
Vr ≈ 0 V
Therefore, the voltage drop across the capacitor is approximately 501 V, while the voltage drop across the resistor is approximately 0 V.
Answer: Voltage drop across the capacitor = 501 V, Voltage drop across the resistor = 0 V
2) The current through the resistor (I) after many time constants can be determined using Ohm's Law:
I = emf / R
Substituting the values:
emf = 501 V
R = 4300 Ω
Calculating the current:
I = (501 V) / (4300 Ω) ≈ 0.1165 A
Therefore, the current through the resistor is approximately 0.1165 Amperes.
Answer: Current through the resistor ≈ 0.1165 A
To solve both questions, we need to understand the behavior of a RC circuit. In an RC circuit, the voltage across the capacitor build-up over time while the voltage across the resistor decreases. The time constant (τ) for an RC circuit is given by the formula τ = R * C.
1) To find the voltage drop across the capacitor and across the resistor after a long time, we need to determine the charge stored in the capacitor. The formula to calculate the charge (Q) on a capacitor is Q = C * V, where C is the capacitance and V is the voltage.
Assuming the circuit has been connected for a long time, the capacitor will be fully charged, and the voltage across the capacitor (Vc) will equal the voltage of the source (emf). Therefore, Vc = 501 V.
The voltage across the resistor (Vr) will be the difference between the emf and the voltage across the capacitor. Since the emf of the source is 501 V and the voltage across the capacitor (Vc) is also 501 V, the voltage across the resistor (Vr) will be 501 V - 501 V = 0 V.
So, the voltage drop across the capacitor is 501 V, and the voltage drop across the resistor is 0 V.
Therefore, the answer is: 501 V, 0 V.
2) To find the current through the resistor (I), we can use Ohm's Law (V = I * R) where V is the voltage drop across the resistor (which we found to be 0 V) and R is the resistance.
Since the voltage drop across the resistor (Vr) is 0 V, the current through the resistor (I) will also be 0 A.
Therefore, the answer is 0 A.