Question

An alternating voltage v(t) = 220 sin 100πt volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is ______.

Options

• 5 ms

• 2.2 ms

• 7.2 ms

• 3.3 ms

Answer 1

An alternating voltage v(t) = 220 sin 100πt volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is 3.3 ms.

Explanation:

Given: The load resistance is R = 50Ω, and the applied voltage is

V(t) = 220 sin (100πt) ..........(i)

To find: t', the time taken for the current to rise from half of the peak value to the peak value.

Current flowing through the alternating current circuit at time t:

I(t) = `V/R = 220/50 sin (100pit)`

I(t) = `22/5 sin (100pit)` .........(ii)

Time required to complete one current cycle:

T = `(2pi)/omega = (2pi)/(100pi) = 1/50 s` ........(iii)

The time needed to attain the current's maximum value:

`t_"peak" = T/4 = 1/200 s` ......(iv)

Let I_peak be the highest possible current via the circuit.

Hence, `I_"peak"/2` will be half the current's peak value.

Let t₁ be the time it takes the current to reach half its maximum value.

As a result of equation (i),

`I_"peak"/2 = I_"peak" sin (100pit_1)`

`100pit_1 = pi/6`

`t_1 = 1/600` ......(v)

The time required for the current to rise from half of the peak value to the peak value is calculated using equations (iv) and (v):

`t^' = 1/200 - 1/600 = 1/300 = 3.3` ms