1.

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

Answer»

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

V = Risin 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.



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