An alternating voltage V(t)=220sin100πt volt is applied to a purely resistive load of 50Ω. The time taken for the current to rise from half of the peak value to the peak value is:

The given alternating voltage is V(t) = 220sin100πt.

The peak voltage, Vm = 220V.

The current in a purely resistive load when an alternating voltage is applied is given by Ohm's law, I = V/R.

So, the peak current, Im = Vm/R = 220/50 = 4.4A.

Half of the peak current is Im/2 = 4.4/2 = 2.2A.

The time taken for the current to rise from half of the peak value to the peak value can be found by setting up the equation for the current and solving for the time.

The current as a function of time is I(t) = Imsin100πt.

We set this equal to half the peak current and solve for t:

2.2 = 4.4sin100πt.

Dividing both sides by 4.4 gives:

0.5 = sin100πt.

The inverse sine of 0.5 is π/6 or 30 degrees, so:

100πt = π/6.

Solving for t gives t = 1/600 seconds.

Therefore, the time taken for the current to rise from half of the peak value to the peak value is 1/600 seconds.