Kirchoff's Loop Law: Solving DC Circuit w/5V Power Supply

[Khuwwa]

The problem:

I have one loop of a larger circuit. It has a DC power supply set to 5 V, a 220 Ohm resistor, a junction, a 100 Ohm resistor, a junction, a 470 Ohm resistor, a junction, then it goes back to the power supply.

The current through the 220 Ohm resistor is 11.71 mA (0.01171 A). The current through the 100 Ohm resistor is 4.72 mA (0.00472 A), and the current through the 470 Ohm resistor is 4.10 mA (0.00410 A). Then there is a junction, and the current going back to the battery is 6.51 mA (0.00651 A).

My attempt at a solution:

I decided to traverse the loop clockwise, which is the direction of the current. I found out that when I traverse a resistor in the same direction as the current, the voltage drop is -(current*resistance). I calculated this for all three resistors, obtaining voltage drops of -2.58 V, -0.472 V, and -1.93 V, for the respective resistors.

I then turned to the power supply. I found out that when traversing a power supply from its negative terminal to its positive terminal, the voltage drop is
-emf. To find emf, I used the equation:

emf = current*load resistance + current*internal resistance.

My question:

How do I find the internal resistance of my power supply? I assume the load resistance is just the combined resistances of the three resistors; is this correct? And what do I use for current, the current flowing into the power supply or the current flowing out of it?

 

[mjsd]

can you show us your circuit diagram? or is that not given?

 

[Khuwwa]

I tried as best I could to describe the circuit in my first paragraph: a DC power supply and three resistors, with varying currents along the circuit because it is only one closed loop of a larger circuit.

 

[mjsd]

ok... I think I kind of get the picture... so I would guess that the internal resistance is taken as in series with the power supply. Now, you need to identify the current in the same branch where the power supply/internal resistance is located, my guess is that it is the 6.51mA. So you will have something like
V(supply) = R_int I_int + R1 I_1 + R2 I_2 + R3 I_3

 

[Khuwwa]

5 volts = r * 0.00651 A + 220 * 0.01171 A + 100 * 0.00472 A + 470 * 0.00410 A

 

[Khuwwa]

I got internal r = 2.76 Ohms, does that seem like a reasonable answer?

I have to put it into emf = IR + Ir now. Is R (load resistance) just 220+100+470? And what about the two Is?

 

[Khuwwa]

I solved the problem, thank-you for your help!