Im having trouble with these practical applications in unit 8. Here is question number 2:

2 resistors are connected in series. one resistor has a resistance of 1000 Ω, and the other resistor value is unknown. the circuit is connected to a 50 volt power supply, and there is a current of 25mA flowing un the circuit. if a 1500Ω resistor is connected in parallel with the unknown resistor, how much current will flow in this circuit?

 

Hmm, value of unknown resistor can be found if you understand series connection.
And from Ohms Low we know
V = 25mA * 1000 = 25V
http://www.allaboutcircuits.com/vol_1/chpt_5/1.html

 

please explain ur equation again. i think im starting to get it.

V = 25mA * 1000 = 25V

so 1 volt = 25 mA? and 25 volts equals 1000 ohms?

 

You have two resistor connect in series.
First R1 resistor is equal R1 = 1000Ω = 1KΩ second Rx is unknown.
Thous two resistor are connect to 50V power supply.
Total resistance of a circuit is equal
Rtot = 50V/25mA = 2KΩ, and voltage on R1 is equal 25V.
So the remain voltage of the supply voltage must be across Rx.
So Rx = ??

 

You have a circuit with three elements connected in series. These three elements are: 50 V supply, 1k resistor, unknown resistor (RX). Current in this circuit is 25 mA. Since all the elements are in series, 25 mA passes through each element. Therefore:
50V = 25mA(1k+RX)
solve this for RX

 

so the answer is 25 volts?

 

Yes 25V will be present across Rx resistor.
So from Ohms Low Rx = 25V/25mA

 

so the answer is 25 volts?

That was step one. You now know all the elements that make your circuit. Using this information you can find current. You will have 50V supply in series with R1=1k resister in series with combination of RX=1k resister in parallel with 1.5k resistor.

Simplify the two resistors in parallel (RX and 1.5k), that will give you one resistor in series with the R1=1k resistor. Simplify two resistors in series, that will give you one resistor. Use Ohm's Law to find current.