Question

An AC voltage source and a resistor are connected in series to make up a simple AC circuit. If the source voltage is given by ΔV = ΔVmax sin(2πft) and the source frequency is 16.9 Hz, at what time t will the current flowing in this circuit be 55.0% of the peak current?

Answer 1

Answer:

t = 5.48 × 10⁻³ s

Explanation:

Given:

ΔV = ΔVmax × sin(2πft)

frequency, f = 16.9Hz

thus,

ΔV = ΔVmax × sin(2π×16.9×ft)

Now,

Let 'R' be the resistance

Also according to the ohms law

i = V/R

where,

i = current

V = voltage

hence,

$i=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}$

also, given at time 't' the current in the circuit is 55.0% of the peak current

thus

$i=\frac{55}{100}\times \frac{\Delta V_{max}}{R}=0.55\times \frac{\Delta V_{max}}{R}$

thus,

$0.55\times \frac{\Delta V_{max}}{R}=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}$

or

$0.55=sin(2\pi \times 16.9\times t)}$

or

$0.5823=(2\pi \times 16.9\times t)}$

or

t = 5.48 × 10⁻³ s (Answer)