# find unkown values in series circuit

Discussion in 'Homework Help' started by the_fundamental_freq, May 7, 2011.

1. ### the_fundamental_freq Thread Starter New Member

May 7, 2011
8
0
hi everyone,

have been given this circuit as revision for an upcoming exam, and really cant remember where too start (may just need a bit of memory jogging). its a series circuit so the '20mA' stated would be the total current? also i think that the way to work it out would be to use p=i^2/r (as 'P6) is given but is this not the total power dissipation?
i realize im a total noob on here and im sorry if i have annoyed anyone by asking for help on my first post! (will try and help others if i can)

many thanks

link to image: http://postimage.org/image/1v373e2mc/

2. ### jegues Well-Known Member

Sep 13, 2010
735
43

You can start with the middle branch. First you should find the voltage drop V2.

After that, you know that R3=R4, so call it R, they are in series so we can combine them as 2R.

Now write a KVL in the loop on the RHS and solve for the required voltage drop across the combined series resistor 2R and thus solve for R.

This should be enough to get you started.

3. ### the_fundamental_freq Thread Starter New Member

May 7, 2011
8
0
ah im with ya i think?
v2= 2V
R 3&4 are 285 Ohms
V 3&4 5.7
R5=330ohms
thus (excluding R1&R6) Rt=1kOhm and and tot V drop 20v

would i then just follow the same principle for R1 & R6 ie combine to make 2R

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,082
1,149
V2 is equal 2V
V3 = V4 = 5.7V OKi
R5 = 330Ω

As for R6, you know the power P = V * I = 0.112W
But we also know the Ohm's law
I = V/R and V = I*R so
P = V*V/R = V^2 *R = I*R * I = I^2 * R
Form this equation we can solve for R
R = ?

Last edited: May 7, 2011
5. ### the_fundamental_freq Thread Starter New Member

May 7, 2011
8
0
i^2/p=280 ohms?

May 7, 2011
8
0
then
v1=5.6
r6= 220
v6= 4.4?

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,082
1,149
P = I^2 * R
I divided by I^2

$\frac{P}{I^2} =\frac{I^2}{I^2} * R$

$\frac{P}{I^2} = R$

$R = \frac{P}{I^2}=\frac{112mW}{20mA^2} = 280\Omega$

V6 = I * R = 20mA * 280Ω = 5.6V

V1 = 30V - 20V - 5.6V = 4.4V

R1 = V/I = 4.4V/20mA = 220Ω

8. ### the_fundamental_freq Thread Starter New Member

May 7, 2011
8
0
cool thanks for that, so i basically got the values mixed up from R1 & R6, how do you know which to do first?

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,082
1,149
Simply you need experience to be able to solve this type of problems.

10. ### the_fundamental_freq Thread Starter New Member

May 7, 2011
8
0
if there is no way to determine which to do first, is there anyway of checking after?

11. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,819
1,736
Checking after is the easy part. Since you should now have all the resistor values of this simple series circuit then you sum them, divide the 30V source by the resistance, and see if you get the 20 mA back.

If you don't, try, try again.