A 20 nF capacitor is connected across an AC generator that produces a peak voltage of 5.0 V. a. At what frequency f is the peak current 50 mA? b. What is the instantaneous value of the emf at the instant when iC=IC?

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ramirezhereva · Accepted Answer

a) The current peak through the capacitor is defined as a:  IC=ϵ0XC  IC=ϵ012⋅π⋅f⋅C  IC=2⋅π⋅f⋅C⋅ϵ0  If we express the frequency and substitite, we can compute the value for the frequency:  f=IC2⋅π⋅C⋅ϵ0  f=50 mA2⋅π⋅20⋅10−9 F⋅5 V  f=79.6 kHz  b) At he capacitor, the current always leads by the 90∘. So if we present the current in an instantaneous value, the current will be:  ic=IC⋅cos⁡(ω⋅t+π2)  in a moment when the iC=IC, equality must apply:  cos⁡(ω⋅t+π2)=1  ω⋅t+π2=2⋅n⋅π  ω⋅t=2⋅n⋅π−π2  As the voltage delay by π2 in respect to current, the voltage is given as:  vc=ϵ0⋅cos⁡(ω⋅t+π2−π2)  vc=ϵ0⋅cos⁡(ω⋅t)  vc=ϵ0⋅cos⁡(2⋅n⋅π−π2)  vc=ϵ0⋅cos⁡(−π2)  vc=ϵ0⋅0  vc=0

A 20 nF capacitor is connected across an AC generator

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