Question #226818

We keep the switch at position x for a sufficient time and thereafter (at t0) connect it to y, in the cercuit shown below : what amount will be measured through the resistor after 5.0 micro-seconds from t0, if v=9v, R= 10ohms and c=1 micro F

Expert's answer

Given:

R=10 ΩR=10\:\Omega

C=1∗10−6 FC=1*10^{-6}\:\rm F

V=9 VV=9\:\rm V

t=5∗10−6 st=5*10^{-6}\:\rm s


The time constant

τ=RC=10∗1∗10−6=10−5 1/s\tau=RC=10*1*10^{-6}=10^{-5}\:\rm 1/s

The current through the resistor

I=VRe−t/τI=\frac{V}{R}e^{-t/\tau}

I=910e−5∗10−6/10−5=0.55 AI=\frac{9}{10}e^{-5*10^{-6}/10^{-5}}=0.55\:\rm A

The voltage

VR=I∗R=0.55∗10=5.5 VV_R=I*R=0.55*10=5.5\:\rm V


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