A battery of 9 V charges a capacitor of 47 ìF through a resistor.
What resistor value needs to be used to obtain a voltage across the
capacitor of 8 V when 10 seconds have elapsed ? You will need to
obtain the standard equation for the voltage across the capacitor
charging in this circuit configuration.
I have done the first part of the question but I'm not sure if that's the correct answer.
10 s. = 10,000 mS.
x = t/RC, Vr = 9/e^x = 9-8 = 1,
e^x = 9, X = 2.197 = 10,000/(R*47), 2.197 =213/R, 2.197R = 213, R = 96.8Kohms.
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To solve this question, we need to use the standard equation for the voltage across a charging capacitor in an RC circuit, which is given by:
Vc = V0 * (1 - e^(-t/RC))
where Vc is the voltage across the capacitor at time t, V0 is the initial voltage across the capacitor (0 V in this case), e is the mathematical constant approximately equal to 2.71828, t is the time elapsed, R is the resistance in ohms, and C is the capacitance in farads.
Given that V0 = 0 V, Vc = 8 V, t = 10 seconds, and C = 47 μF (or 47 x 10^(-6) F), we can rearrange the equation to solve for R:
8 V = 0 V * (1 - e^(-10/RC))
Since the initial voltage is 0 V, this simplifies to:
8 V = (1 - e^(-10/RC))
Now, we can isolate the exponential term:
e^(-10/RC) = 1 - 8 V
Next, let's isolate the exponential term by subtracting 1 from both sides:
e^(-10/RC) - 1 = -7 V
Finally, to solve for R, we need to take the natural logarithm (ln) of both sides:
ln(e^(-10/RC) - 1) = ln(-7 V)
Now, we can solve for R by rearranging the equation:
-10/RC = ln(-7 V)
Dividing both sides by -10:
1/RC = -ln(-7 V) / 10
Finally, we can find the resistor value, R, by rearranging the equation:
R = 1 / (C * -ln(-7 V) / 10)
Now, we can calculate R using the given values:
R = 1 / (47 x 10^(-6) F * -ln(-7 V) / 10)
Please note that you might have made a mistake in solving the equation, and it's a good idea to double-check your work or use a calculator to ensure accuracy.