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