A particular signal source produces an output of 30 mV when loaded by a 100-k oh

Question

A particular signal source produces an output of 30 mV when loaded by a 100-k ohm resistor and 10 mV when loaded by a 10-k ohm resistor. calculate the thevenin voltage, Norton current, and source resistance.

Explanation / Answer

We know the voltage division rule as Vout=Vth(R)/(R+Rs) here R is the load and Rs is the source resistance.. For R=100kohms,Vout=30mV substituting this values in Vout gives 30x10^-3=Vth(100x10^3)/(Rs+(100x10^3)) =>0.003Rs+3000=Vth(100x10^3) for R=10kohms,Vout=10mV =>10x10^-3=Vth(10x10^3)/(Rs+(10x10^3)) =>100+0.001Rs=Vth(10x10^3) multiplying this equation by 10 and equating it to Vth(100x10^3) =>1000+0.01Rs=0.003Rs+3000 =>0.007Rs=2000 =>Rs=285.714kohms substituting Rs in above equation gives 100+0.001(285710)=Vth(10000) =>Vth=38.571mV norton current In is given by In=Vth/Rs=0.135x10^-6 A

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