the ouput of a generator that product a maximum voltage of 9

the ouput of a generator that product a maximum voltage of 90 volt is to be attenuated at 10 volt for input to a filter. specify value of the resistor in an attenuation network such that the loading error of the voltage at the ouput terminal of the generator is 0.1%. the ouput impedance of the generator is 10 ohm and the filter has an impedance of 100k

Solution

Solution :

For the 10 ohms of the generator to be loaded by less than 0.1%, the load has to be more than 10000 ohms, as 0.1% is 0.001, and one over that is 1000. This means 1000 x 10 = 10000 ohms.

That means the current is Voltage/ resistance = 90v/10k = 9mA which causes a E=IR = 9ma*10 = 90mV drop across the 10 ohm resistor, which is 0.1% of 90 volts.

So the Resistors can be:

top resistor is 80V/9mA = 8.8888 kohm
bottom resistor is 10v/9mA = 1.1111 kohm
of course, you can subtract 10 ohms from the 8.8888 kohm and use 8.8788k ohms Because of the error and this value wont have any error.

The value of R2 is determined by (V of R2) / (I of R2)) = 10V / .0089A = 1123.6 ohms.

Hence resistances in the divider consists of a 8.8788k and 1.1236k