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sertanlavr [38]
3 years ago
11

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?
Physics
1 answer:
ra1l [238]3 years ago
8 0

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)

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