There is circuit with one resistor, one capacitor, one switch and a dc voltage source
resistor = 200Ω
dc voltage = 10v
Capacitor = 10 uf = 0.00001f
can someone help me on question 2 and 3
1. what is the RC time constant of the circuit?
answer: (200Ω)(0.00001f)=0.002s
2. What is the time from the instant that the switch closes until the capacitor fully charges? Assume that the capacitor is fully discharged prior to the switch closing?
3. What is the current in the circuit the instant after the switch closes? Assume that the capacitor is fully discharged prior to the switch closing
2 --- forever in theory e^-t/RC never actually reaches zero
3 --- at first all the electrons try to go to the empty go to the capacitor but the voltage on the capacitor is zero (no charge yet) so the 10 volts is across the resistor
i = V/R = 10 volts / 200 ohms at the start
2 usually, in practice, many consider 5 time constants will fully charge from zero in an RC circuit.
@Bobpursley, do you mean t=5*rc?
Yes, what bobpursley said means t=5*rc, so anonymous you are correct.
To answer question 2 and 3, we need to understand the behavior of a RC circuit when a switch is closed.
In an RC circuit, when the switch is initially closed, the capacitor starts to charge up over time. The time it takes for the capacitor to fully charge is determined by the RC time constant, which we have already calculated in question 1.
The RC time constant (τ) is the product of the resistance (R) and the capacitance (C), given by τ = R * C.
For this circuit, the RC time constant is (200Ω)(0.00001F) = 0.002s (as you have correctly calculated).
Now let's move to question 2:
2. What is the time from the instant that the switch closes until the capacitor fully charges? Assume that the capacitor is fully discharged prior to the switch closing.
To determine the time taken for the capacitor to fully charge, we need to know the behavior of the charge on the capacitor over time. In an RC circuit, the charge on the capacitor follows an exponential growth curve given by the equation:
Q(t) = Q0 * (1 - e^(-t/τ))
Where:
- Q(t) is the charge on the capacitor at time t
- Q0 is the final charge on the capacitor (when it is fully charged)
- e is the mathematical constant approximately equal to 2.71828
- t is the time in seconds
- τ represents the RC time constant (0.002s in this case)
The time it takes for the capacitor to fully charge (t_charge) is when Q(t_charge) = Q0, which means the charge on the capacitor reaches its maximum value.
In this case, since the capacitor is fully discharged prior to the switch closing, the maximum charge Q0 is equal to the total charge that the capacitor can hold, given by Q0 = C * V, where V is the applied voltage.
For this circuit, the capacitance C is 0.00001F and the voltage V is 10V, so Q0 = (0.00001F)(10V) = 0.0001C.
Now we can solve for t_charge using the equation Q(t) = Q0 * (1 - e^(-t/τ)):
0.0001 = 0.0001 * (1 - e^(-t_charge/0.002))
Dividing both sides by 0.0001, we get:
1 = 1 - e^(-t_charge/0.002)
Subtracting 1 from both sides gives:
0 = -e^(-t_charge/0.002)
Taking the natural logarithm (ln) of both sides gives:
ln(0) = -t_charge/0.002
The natural logarithm of 0 is undefined, so this equation cannot be solved. This means that the capacitor never fully charges in this scenario. As time approaches infinity, the charge on the capacitor gets closer and closer to full, but it can never actually reach it.
Therefore, we cannot provide a specific time when the capacitor fully charges in this case. The time it takes for the charge to approach but never reach the maximum value is infinite.
Moving on to question 3:
3. What is the current in the circuit the instant after the switch closes? Assume that the capacitor is fully discharged prior to the switch closing.
Immediately after the switch is closed, the capacitor behaves as an open circuit because it is fully discharged. In this case, the current in the circuit is determined by Ohm's law, which states that current (I) is equal to the voltage (V) divided by the resistance (R).
For this circuit, the resistance is 200Ω and the voltage is 10V. Therefore, the current (I) is given by:
I = V / R = 10V / 200Ω = 0.05A
Thus, the current in the circuit, at the instant after the switch closes, is 0.05A or 50mA.
To summarize:
2. Since the capacitor never fully charges, the time it takes for the capacitor to fully charge is infinite.
3. The current in the circuit the instant after the switch closes is 0.05A or 50mA.