The voltage (V) applied to a 0.25 F capacitor is given by the equation :
v= (t^2-2t)^3/5t
Find the current at t = 3.7 s. (i=C dv/dt)
Round to 3 significant figures
To find the current at t = 3.7 s, we will use the equation i = C(dv/dt), where i represents the current, C represents the capacitance, and dv/dt represents the rate of change of voltage with respect to time.
Given: The voltage equation v = (t^2 - 2t)^3 / 5t and C = 0.25 F
To find dv/dt, we need to differentiate the voltage equation with respect to time. Let's perform the differentiation step by step:
Step 1: Expand the equation
v = (t^2 - 2t)^3 / 5t
= (t^2 - 2t)(t^2 - 2t)(t^2 - 2t) / 5t
Step 2: Differentiate each term using the product rule
dv/dt = (d/dt) [(t^2 - 2t)(t^2 - 2t)(t^2 - 2t) / 5t]
= (d/dt) [(t^2 - 2t)^3 / 5t]
= [(d/dt)(t^2 - 2t)^3 * 5t - (d/dt)(5t)(t^2 - 2t)^3] / (5t)^2
Step 3: Differentiate each term using the chain rule and the power rule
(d/dt)(t^2 - 2t)^3 = 3(t^2 - 2t)^2 * (d/dt)(t^2 - 2t)
= 3(t^2 - 2t)^2 * (2t - 2)
= 6(t^2 - 2t)^2 * (t - 1)
Step 4: Substitute the differentiated term back into the equation
dv/dt = [6(t^2 - 2t)^2 * (t - 1) * 5t - 5(t^2 - 2t)^3] / (5t)^2
= [30t(t^2 - 2t)^2 * (t - 1) - 5(t^2 - 2t)^3] / (5t)^2
Now that we have the expression for dv/dt, we can find the current at t = 3.7 s by substituting the value into the expression.
i = C(dv/dt)
= 0.25 * [(30t(t^2 - 2t)^2 * (t - 1) - 5(t^2 - 2t)^3) / (5t)^2] -- substitute C = 0.25 F and dv/dt expression
Now, we substitute t = 3.7 s into the equation and round the answer to 3 significant figures:
i = 0.25 * [(30 * 3.7(3.7^2 - 2(3.7))^2 * (3.7 - 1) - 5(3.7^2 - 2(3.7))^3) / (5 * 3.7)^2]
Calculating this expression will give you the current at t = 3.7 s, rounded to 3 significant figures.