Engineering Electrical Engineering Circuits

[Mark Nussbaum]

Homework Statement

Homework Equations

V=IR
kirchoff's Voltage/current law

The Attempt at a Solution

Going off the answer i got correct, 65V / 138.1788Ohms (Req of total circuit) for a total current of .47A. I think this is where i messed up but don't know how to continue...

 

[berkeman]

Homework Statement

View attachment 211115

Homework Equations

V=IR
kirchoff's Voltage/current law

The Attempt at a Solution

Going off the answer i got correct, 65V / 138.1788Ohms (Req of total circuit) for a total current of .47A. I think this is where i messed up but don't know how to continue...

Can you describe the series & parallel combinations you used to the right of the dashed line to get the equivalent resistance?

 

[magoo]

You really need to show the steps that you took to get the 4 answers. Otherwise, we have no idea if you're on the right track on not.

 

[Mark Nussbaum]

Can you describe the series & parallel combinations you used to the right of the dashed line to get the equivalent resistance?

For Req to the right of the dashed line,
R3+R4=385 Ohms
1/(1/R2+1/385) = 91.4 Ohms
91.4 Ohms + 70 Ohms = 161.4 Ohms

This is in parallel with the 50 Ohms resistor so i combined them with
1/(1/50+1/161.4)=38.17 Ohms
then adding in series with the 100 Ohms a total of 138.17 Ohms.

Using the total resistance I divided the voltage source to find the current,
65V/138.17 Ohms =.47A

After that I calculated the voltage drop across the 100 Ohms resistor V=IR
V=.47 A*100 Ohms= 47 Volts

Since the other resistor is in parallel with the Req on the right side of the dashed line I assumed the same voltage would go to each branch
65-47=18V

After that I calculated the current that went through the 50 Ohm resistor
18V/50 Ohms= .36A
so the remaining would run through the other branch

Once again calculating voltage drop across R1
V=.11 A * 70 Ohms = 7.7 V

and since R2 is in parallel with R3/4 I assumed this voltage would V2

 

[magoo]

Your calc for Req is correct.

When you put that in parallel with the 50 ohms and get the series resistance, that too is correct. However, I get 0.4704 A.

The drop thru the 100 ohm resistor is 47.04 V (or 47 in your case). This leaves 17.96 V for V2, which is close enough to the 18 V you came up with.

The next step is to continue the calculation based on 17.96 V and the equivalent resistance to the right of V2 (161.485 ohms).

This gives you a current of 0.111 A, leading to a voltage drop across the 70 ohm resistor of 7.79 V. I think this is what you put for V2, when it is really the drop across that resistor. The resulting V2 is 17.96 – 7.79 or 10.17 V.

This will naturally change your value for i4.