Answer:
[tex]V(t)= 240V* H(t-5)[/tex]
Explanation:
The heaviside function is defined as:
[tex]H(t) =1 \quad t\geq 0\\H(t) =0 \quad t <0[/tex]
so we see that the Heaviside function "switches on" when[tex]t=0[/tex], and remains switched on when [tex]t>0[/tex]
If we want our heaviside function to switch on when [tex]t=5[/tex], we need the argument to the heaviside function to be 0 when [tex]t=5[/tex]
Thus we define a function f:
[tex]f(t) = H(t-5)[/tex]
The [tex]-5[/tex] term inside the heaviside function makes sure to displace the function 5 units to the right.
Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ([tex]H(t-5) =1[/tex] when [tex]t\geq 5[/tex], so it becomes just a 1, which we can safely ignore.)
Therefore our final result is:
[tex]V(t)= 240V* H(t-5)[/tex]
I have made a sketch for you, and added it as attachment.
