- [Teacher] Suppose a circuit contains
an electromotive force, a battery,
that produces a voltage of E of t volts,
a capacitor with a capacitance of C farads
and a resistor with a resistance of R ohms.
The voltage drop across the capacitor is Q divided by C
where Q is the charge in coulombs
so in this case Kirchoff's Law gives
R times I equals Q divided by C equals E of t.
And since I is equal to dQ dt,
we have the differential equation
R times dQ dt plus Q divided by C
or one over C times Q equals E of t.
Now suppose the resistance is ten ohms,
the capacitance is 0.05 farads,
a battery gives a constant voltage of 60 volts,
and the initial charge is Q of zero equals zero coulombs.
I'm going to find the charge and the current at time t.
So I've listed all the given information here below.
We're given C, R, E of t, and Q of zero equals zero.
We want to find the charge function
and the current function.
To find these, we'll be solving the differential equation
given here for Q and then we can find I
by determining dQ dt.
Let's set this up on the next slide.
We'll first substitute all of the known values
into the differential equation.
So we have R which is ten times dQ dt
plus one over C would be one over 0.05
times Q which is unknown equals E of t
which is the constant 60.
We can solve this differential equation
by using separation of variables,
but let's first simply this term here.
One over five hundredths or one over one twentieth
is equal to 20, so this term is really 20 Q.
So we have ten dQ dt plus 20 Q equals 60.
Let's go ahead and solve this equation for dQ dt.
So I'll subtract 20 Q on both sides.
So we have ten times dQ dt equals 60 minus 20 Q.
Let's divide both sides by ten so we have dQ dt
equals six minus two Q.
And now let's multiply both sides by dt,
so we'd have differential Q
equals the quantity six minus two Q times dt.
And now we want this function of Q on the left.
Let's multiply both sides by one over
the quantity six minus two Q.
So this will give us one over the quantity
six minus two Q dQ equals dt.
Now that we have all the Qs on the left
and the ts on the right,
we'll integrate both sides of the equation.
So we'd have the integral of one
over the quantity six minus two Q dQ
equals the integral of dt or one dt.
We'll have to perform u substitution here on the left
where we'd let u equal six minus two Q
and therefore differential u is equal to negative two dQ.
So if we divide both sides by negative two,
notice how we'd have negative one half du equals dQ.
So if it's helpful, we can think of all of this
as negative one half times one over u du.
So the integral in terms of u would be negative one half
natural log absolute value of u
which means in terms of Q we'd have
negative one half times the natural log
of the absolute value of six minus two Q.
We'll put the constant of integration on the right,
so this is equal to the integral of
one to the respect of t is t,
so we have t plus c sub one.
Now we need to work on solving this for Q.
Let's multiply both sides by negative two.
So we'd have natural log absolute value of six minus two Q
equals negative two times the quantity t plus c sub one.
So we have natural log absolute value
of six minus two Q equals negative two t
and then negative two times c sub one
is just another constant.
Let's let negative two times t sub one
be equal to c sub two.
So we have plus c sub two.
Now let's write this log equation
as an exponential equation.
We know natural log is base e,
which means e raised to power of negative two t
plus c sub two must equal the quantity six minus two Q.
So let's write that as six minus two Q equals
e raised to the power of negative two t plus c sub two.
Let's continue on the next slide.
Because we're adding exponents here,
we can write this as a product.
We can write the right side as e to the negative two t
times e raised to the power of c sub two.
But e raised to the power of c sub two is just a constant.
Let's let this equal c sub three.
So we can write this as six minus two Q
equals c sub three times e
raised to the power of negative two t.
And now to solve for Q, we'll subtract six on both sides.
So we have negative two Q equals c sub three
times e raised to the power of negative two t minus six
and then I'll divide both sides by negative two.
So we can write this as Q, or Q of t
equals, let's write this as negative one half c sub three
e to the negative two t,
and then we'd have plus three.
Notice negative six divided by
negative two is positive three.
But negative one half times c sub three
is just another constant.
Let's let c be equal to negative one half c sub three
and write this as Q of t equals c times e
raised to the power of negative two t plus three.
So this is the general solution
to the differential equation,
but we want to find the particular solution,
and we can find the particular solution
because we were given that Q of zero is equal to zero.
So if we know that Q of zero is equal to zero,
we can solve for c by substituting zero for t
and zero for Q of t.
That would give us the equation zero equals
c times e raised to the power
of negative two times zero, plus three.
Well this is e to the zero which is equal to one,
so we just have zero equals c plus three.
So c equals negative three.
So the particular solution is Q of t
equals negative three times e
raised to the power of negative two t plus three.
So going back to our very first slide,
we now have the charge function Q of t
equals negative three times e
raised to the power of negative two t, plus three.
And now that we have Q of t,
we can find I of t because I of t is equal to dQ dt.
So again we have Q of t.
And we were given that I of t equals dQ dt
so I of t is equal to the derivative
of negative three times e to the negative two t plus three
with respect to t,
so we'd have negative three times e to the negative two t
times negative two.
And the derivative of three is zero,
so I of t is equal to positive six e
raised to the power of negative two t.
So now we also have the current function I of t.
I of t equals six e raised to the power of negative two t.
I hope you found this helpful.