Since the voltage dropped from 10V10 V10V to 8V8 V8V , therefore, the voltage 2V2 V2V at the new load, therefore the current at the load I=URR=2104=0.0002I =\frac {U _ {R}} {R} =\frac {2} {10 ^ 4} = 0.0002I=RUR=1042=0.0002 A.
Therefore, resistor resistance Ramplifier=UoutputI=80.0002=40R_{amplifier}=\frac{U_{output}}{I}=\frac{8}{0.0002}=40Ramplifier=IUoutput=0.00028=40 KΩ\OmegaΩ.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments