There is an electric circuit with a supplied voltage (electromotive force)E,a capacitor with capacitance C, and a resistor with resistance R.When a switch s in the circuit is opened at time t=0 the current I through the circuit begins to decrease exponentially as a function of time t (measured in seconds after the switch is opened) given by
$I=\frac{E}{R}e^{\frac{-t}{RC}}$
for $t\geq0$
A) I want to sketch a rough graph of I as a function of t.
B)Note that at time $t=0$ the current is $I=\frac{E}{R}$ (measured in amperes) which is the familiar formula from Ohm's Law. That is the peak value of I. What is the current I at time t= 5RC? This answer should be in the decimal form as a percentage of peak current $\frac{E}{R}$ (e.g. $0.42\frac{E}{R}$ means 42% of the peak current.
C) I want to find the average current in the circuit over the time interval$[0,5RC]$. This answer should also be in decimal form as a percentage of peak current.
Solution. How to solve these questions? I know the average value of function over a closed interval [a,b] is given by
$f=\frac{1}{b-a}\displaystyle\int_a^b f(x)dx$
A) and B) should be pretty easy.
For A) you need to draw the graph of $y = ce^{-kx}$; where $c > 0$ and $k > 0$ are constants. So a graph "similar" to $e^{-x}$ will do.
For B) you just need to substitute $t = 5RC$ in the equation given, just need a calculator for the decimal form asked.
For C) calculate $\frac{\int_0^{5RC} I dt}{5RC}$ as you already figured out. Integral of the exponential function should be easy too