A 15.2-F capacitor has an initial voltage of 2.00 V. It is charged with a current given by i=t�ã5=t^2. Find the voltage across the capacitor at 1.75 seconds.
V = Q/C = Initial Voltage + (1/C)integral of i dt
Is the answer 2.25 v
To find the voltage across the capacitor at 1.75 seconds, we need to integrate the current function over the given time interval.
The current function is given by i = t^2. We can find the charge flowing into the capacitor by integrating the current function with respect to time:
Q = ∫(i) dt = ∫(t^2) dt
Integrating t^2 with respect to t gives:
Q = (1/3)t^3 + C
We can apply the initial condition, which states that the initial voltage of the capacitor is 2.00 V. The voltage across a capacitor is given by the equation:
V = Q/C
where Q is the charge stored on the capacitor and C is the capacitance. We can rearrange this equation to solve for Q:
Q = VC
Substituting the expression for Q obtained from the integral:
(1/3)t^3 + C = VC
Since the initial voltage of the capacitor is 2.00 V, and the capacitance is given as 15.2 F, we can substitute these values into the equation:
(1/3)t^3 + C = (2.00 V)(15.2 F)
Simplifying this equation, we have:
(1/3)t^3 + C = 30.4 V∙F
To find the constant of integration, C, we can use the initial condition. At t = 0, the charge on the capacitor is zero, so we have:
(1/3)(0)^3 + C = 0
Simplifying, we find:
C = 0
Substituting C = 0 back into the equation, we have:
(1/3)t^3 = 30.4 V∙F
Multiplying both sides by 3:
t^3 = 91.2 V∙F
Now, we need to solve for t. Taking the cube root of both sides:
t = ∛(91.2 V∙F)
Evaluating this expression, we find:
t ≈ 4.517 seconds
Thus, the time taken to charge the capacitor to a voltage of 2.00 V is approximately 4.517 seconds.
To find the voltage across the capacitor at 1.75 seconds, we can substitute this value into the expression for voltage:
V = (1/3)(1.75)^3
Evaluating this expression, we find:
V ≈ 3.245 V
Therefore, the voltage across the capacitor at 1.75 seconds is approximately 3.245 volts.