T=RC, T= time in seconds , R= Resistance in ohms , C = capacitance in farads. Note, that it takes approximately 5 time constants to fully charge or discharge a capacitor. T=(30000)(0.0001) >>> T=3X 5 = 15 SECS for total charge/discharge.
With the two 10 kOhm resistors it would take 10 seconds for the capacitor to fully charge/discharge. Adding a third 10kOhm resistor would change the time to 15 seconds to fully charge/discharge. This is based on calculating the time constants to be t=2 and t=3 respectively, and it taking 5 time constants for a capacitor to fully charge/discharge.
If an additional 10 kOhm resistor is added into the circuit, the time constant takes to charge the capacitor is 18 seconds; it also takes the same time constant (18 seconds) to discharge the capacitor.
So is it to say that if you complete a circuit board’s circuit with a jumper cable, any resistors on the board would discharge any existing capacitors?
Well that depends on what the circuit is doing and where you place the jumpers. If you use a jumper directly on the capacitor then you are bypassing the resistors on the circuit board (which is usually how its done), so you would have a very low resistance (just the resistance of the jumper) so you would discharge very quickly
if it takes 5 time constants for a capacitor to fully charge and discharge we do simple math and come to the conclusion that with 2 resistors the time it takes to charge and discharge is 10seconds. divide that by 5 and you have a time constant of 2 seconds. with one resistor, it takes 5 seconds. divide by 5 and you have a time constant of 1 second. assuming all is wonderful and equal, adding a 3rd resistor will add another 5 seconds to the total time to charge and discharge which would be 15 seconds. divide by 5, and you have a time constant of 3 seconds.
Man, when he dropped the battery wires, they looked like they short circuited. ALWAYS disconnect one battery out of the socket so that the two leads don't accidentally touch and burn those batteries quick.
T=RC, T= time in seconds , R= Resistance in ohms , C = capacitance in farads. Note, that it takes approximately 5 time constants to fully charge or discharge a capacitor.
T=(30000)(0.0001) >>> T=3X 5 = 15 SECS for total charge/discharge.
With the two 10 kOhm resistors it would take 10 seconds for the capacitor to fully charge/discharge. Adding a third 10kOhm resistor would change the time to 15 seconds to fully charge/discharge. This is based on calculating the time constants to be t=2 and t=3 respectively, and it taking 5 time constants for a capacitor to fully charge/discharge.
I just think this is really cool... I love physics.
If an additional 10 kOhm resistor is added into the circuit, the time constant takes to charge the capacitor is 18 seconds; it also takes the same time constant (18 seconds) to discharge the capacitor.
So is it to say that if you complete a circuit board’s circuit with a jumper cable, any resistors on the board would discharge any existing capacitors?
Well that depends on what the circuit is doing and where you place the jumpers. If you use a jumper directly on the capacitor then you are bypassing the resistors on the circuit board (which is usually how its done), so you would have a very low resistance (just the resistance of the jumper) so you would discharge very quickly
The time constant goes from 2 seconds to 3 seconds, making the full charge/discharge time 15 seconds
if it takes 5 time constants for a capacitor to fully charge and discharge we do simple math and come to the conclusion that with 2 resistors the time it takes to charge and discharge is 10seconds. divide that by 5 and you have a time constant of 2 seconds. with one resistor, it takes 5 seconds. divide by 5 and you have a time constant of 1 second. assuming all is wonderful and equal, adding a 3rd resistor will add another 5 seconds to the total time to charge and discharge which would be 15 seconds. divide by 5, and you have a time constant of 3 seconds.
takes about 15/16 seconds to charge and discharge.
I think the time will increase to discharge
That would add another 5 seconds on top of the 10 seconds, so you're total time that it takes to drop would be 15 seconds.
I see
Man, when he dropped the battery wires, they looked like they short circuited. ALWAYS disconnect one battery out of the socket so that the two leads don't accidentally touch and burn those batteries quick.