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
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?
answer: t=5*(rc)
t= 5*(0.002) = 0.01s = 1ms
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.
answer: i=v/r = 10v/200Ω = 0.05A
are my answers correct?
If the circuit is a series RC, all are correct.
it is in a series rc circuit
Yes, your answers are correct.
To calculate the RC time constant of the circuit, you multiply the resistance (R) by the capacitance (C). In this case, R is 200Ω and C is 0.00001F. Multiplying these values together gives you a time constant of 0.002 seconds.
The time it takes for the capacitor to fully charge is typically considered to be around 5 times the RC time constant. In this case, multiplying 5 by 0.002 seconds gives you a total time of 0.01 seconds, which is equivalent to 10 milliseconds (ms).
The current through the circuit the instant after the switch closes can be determined using Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R). In this case, the voltage is 10V and the resistance is 200Ω, so dividing 10V by 200Ω gives you a current of 0.05A (amperes).