Respuesta :
We are given with the equationL
V(t) = 320 e^(-3.1t)
At V = 200
200 = 320 e^(-3.1t)
t = 0.15 s
100 = 320 e^(-3.1t)
t = 0.36 s
It takes 0.36 - 0.15 = 0.21 s for the voltage to drop from 200 to 100 volts.
V(t) = 320 e^(-3.1t)
At V = 200
200 = 320 e^(-3.1t)
t = 0.15 s
100 = 320 e^(-3.1t)
t = 0.36 s
It takes 0.36 - 0.15 = 0.21 s for the voltage to drop from 200 to 100 volts.
Hello there.
How long does it take for the potential to drop from 200 volts to 100 volts?
0.36 - 0.15 = 0.21 s
How long does it take for the potential to drop from 200 volts to 100 volts?
0.36 - 0.15 = 0.21 s