The voltage across the plates of a capacitor at any time t seconds is given by V=Ve -t/CR, where V,C and r are constants. Given V = 300 Volts, C =0.12*10-6 F and R = 4*10 6 ohms
Find the initial rate of change of voltage
and the rate of change of voltage after 0.5 seconds
I think you probably mean
V = Vo e^(-t/RC)
where Vo is 300 volts and the chrge is discharged through the resistor
here 1/RC = 1/(.12*4) = 2.08
so
V = 300 e^-2.08 t
dV/dt = -2.08(300) e^-2.08 t
= -625 e^-2.08 t volts/second
obviously when t = 0 this is -625
you can put in .5 or anything else for t
Well, here's a shocking calculation for you! Let's find the initial rate of change of voltage.
To do this, we need to differentiate the equation V = Ve^(-t/CR) with respect to time (t).
Taking the derivative, we get dV/dt = (Ve^(-t/CR) * (1/CR)).
Now, let's plug in the values given: V = 300 Volts, C = 0.12*10^(-6) F, and R = 4*10^6 ohms.
Substituting those values, we have dV/dt = (300e^(-t/((0.12*10^(-6))(4*10^6)))) / ((0.12*10^(-6))(4*10^6)).
I apologize for all the numbers, but math has a real power to electrify any conversation!
To find the initial rate of change of voltage, we set t = 0 seconds, since we want to know the rate at the very beginning.
Substituting t = 0 seconds into the equation, we get dV/dt = (300e^0) / ((0.12*10^(-6))(4*10^6)).
Simplifying that expression, we have dV/dt = 300 / ((0.12*10^(-6))(4*10^6)).
Now, you can plug in the numbers and calculate the shockingly hilarious initial rate of change of voltage!
As for the rate of change of voltage after 0.5 seconds, you can repeat a similar process. Simply substitute t = 0.5 seconds into the equation we derived earlier and calculate it for another witty result!
Remember, humor sparks interest in learning, so don't be shocked by all these calculations!
To find the initial rate of change of voltage, we need to differentiate the equation V = Ve^(-t/CR) with respect to t.
Differentiating both sides, we get:
dV/dt = -Ve^(-t/CR)/CR
Given V = 300 volts, C = 0.12 * 10^(-6) F, and R = 4 * 10^6 ohms, we can substitute these values into the equation:
dV/dt = -(300e^(-t/(0.12*10^(-6)*4*10^6)))/(0.12*10^(-6)*4*10^6)
To find the rate of change of voltage after 0.5 seconds, we substitute t = 0.5 seconds into the equation:
V = 300e^(-0.5/(0.12*10^(-6)*4*10^6))
Now we can calculate the initial rate of change of voltage and the rate of change of voltage after 0.5 seconds.
To find the initial rate of change of voltage, we need to take the derivative of the voltage equation with respect to time (t).
Given: V = Ve^(-t/CR)
Let's find the derivative of V with respect to t using the chain rule and the exponential derivative rule:
dV/dt = -Ve^(-t/CR) * (-1/CR)
Since V, C, and R are constants, we can substitute their values into the equation:
V = 300 Volts
C = 0.12 * 10^(-6) F
R = 4 * 10^6 ohms
Substituting these values into the equation:
dV/dt = -300 * e^(-t/(4*10^6 * 0.12 * 10^(-6)))
Now, let's substitute t = 0 to find the initial rate of change of voltage:
dV/dt |_(t=0) = -300 * e^(-0/(4*10^6 * 0.12 * 10^(-6)))
Since e^0 is equal to 1, the equation simplifies to:
dV/dt |_(t=0) = -300
Therefore, the initial rate of change of voltage is -300 Volts per second.
To find the rate of change of voltage after 0.5 seconds, we can substitute t = 0.5 into the derivative equation:
dV/dt |_(t=0.5) = -300 * e^(-0.5/(4*10^6 * 0.12 * 10^(-6)))
Using a calculator, we can evaluate this equation to find the rate of change of voltage after 0.5 seconds.