Question

The peak value of AC through a resistor of 10 Ω is 10 mA. What is the voltage across the resistor at time

Answer 1

Data : R = 10 Ω, i₀ = 10 mA, i = \(\cfrac T8\)  

V = Ri₀ sin ωt = (10)(10) sin \(\left(\cfrac{2\pi}T.\cfrac{T}8\right)\)

= 100 sin \(\left(\cfrac{\pi}4\right)\) = \(\cfrac{100}{\sqrt2}\) = 70.72 mV

This is the required voltage.