I can't do your homework,
From part A you know that the resistors must have equal value, lets say R
From B you know the voltages in the circuit. Try to solve the currents ( Kirchoffs law ) as a function of R and finally get the value of R.
Alternatively make the voltage divider equation R series with R parallel with 500ohms, but it is more complicated to solve R from that.

Ok thanks

 

hi yes,
How did you calculate that two 1k's would give 6V at their junction.?
eg: why not two 2K's or any other value providing that the resistor values were the same.??

Show us how you calculated the resistor values for Part A of your question.??

E

 

As long as the two resistors are of equal value and are in series the voltage will remain the same.
I x R = V V/R = I V/I = R

 

As long as the two resistors are of equal value and are in series the voltage will remain the same.
I x R = V V/R = I V/I = R

This true for Part A of the question, but say two, 2K resistors were used to give 6v at the junction.

How could you get the required 3V at the junction, for Part B, with a 500 ohm load.?????

The resistor values for Part A must be calculated correctly, so that the 3v condition for Part B can be achieved.
Look at this image for a two, 2K divider, Part B voltage is only 2v.!

Do you follow OK.?

E

 

This true for Part A of the question, but say two, 2K resistors were used to give 6v at the junction.

How could you get the required 3V at the junction, for Part B, with a 500 ohm load.?????

The resistor values for Part A must be calculated correctly, so that the 3v condition for Part B can be achieved.
Look at this image for a two, 2K divider, Part B voltage is only 2v.!

Do you follow OK.?

E

This true for Part A of the question, but say two, 2K resistors were used to give 6v at the junction.

How could you get the required 3V at the junction, for Part B, with a 500 ohm load.?????

The resistor values for Part A must be calculated correctly, so that the 3v condition for Part B can be achieved.
Look at this image for a two, 2K divider, Part B voltage is only 2v.!

Do you follow OK.?

E