Electronics question

[ChurchOfSubgenius]

Is it possible to calculate R,I or P when you only know V in a series circuit? My teacher gave us a series circuit problem like this:

Vt=120V
V1=12V
V2=18V
V3=90V

and asked us to calculate I2 and P2....this wouldn't be the first time he gave us bad information or asked us something we have yet to go over but I just wanted to do a double check.

The only answer I could think of was: I2= 18V/15% of Rt (similar for P2)

EDIT: Cliff notes..can you calculate resistance from only knowing voltage drops?

 

[Gibson486]

yes, you can. It just splits the voltage up three ways.

Think of 2 equal in series with a 20 V supply. You know right away that the node between the two resistors is 1/2 the voltage of the supply (10V and R1 = R2). In your problem, you know the last resistor is 3/4 the voltage....therefore you know that the resistance must be in terms of just variables ( R3/(R1+R2+R3) = 3/4 ). Also remember that you are saying that this is a series circuit. Remember what happens to current in a series circuit (KCL....very important)?

In the end, you are gonna have 3 equations, with three unknowns. You could solve it linearly, or just use matrix algebra.

 

[ChurchOfSubgenius]

Originally posted by: Gibson486
yes, you can. It just splits the voltage up three ways.

Think of 2 equal in series with a 20 V supply. You know right away that the node between the two resistors is 1/2 the voltage of the supply (10V and R1 = R2). In your problem, you know the last resistor is 3/4 the voltage....therefore you know that the resistance must be in terms of just variables ( R3/(R1+R2+R3) = 3/4 ). Also remember that you are saying that this is a series circuit. Remember what happens to current in a series circuit (KCL....very important)?

In the end, you are gonna have 3 equations, with three unknowns. You could solve it linearly, or just use matrix algebra.

Hmm, seems to me that still leaves me with variables instead of answers, I understand the current is constant (equal) but not how to calculate it exactly without more information.

 

[Quasmo]

If you use this you'll be irresistanceable.

 

[ChurchOfSubgenius]

Originally posted by: Quasmo
If you use this you'll be irresistanceable.

Squeeze a small amount of product onto wet puff. Work it into a lather, massage over your body, and rinse.

Any man who knows what a "wet puff" is will be immediatly excused from the gender.

 

[Gibson486]

Originally posted by: ChurchOfSubgenius

Originally posted by: Gibson486
yes, you can. It just splits the voltage up three ways.

Think of 2 equal in series with a 20 V supply. You know right away that the node between the two resistors is 1/2 the voltage of the supply (10V and R1 = R2). In your problem, you know the last resistor is 3/4 the voltage....therefore you know that the resistance must be in terms of just variables ( R3/(R1+R2+R3) = 3/4 ). Also remember that you are saying that this is a series circuit. Remember what happens to current in a series circuit (KCL....very important)?

In the end, you are gonna have 3 equations, with three unknowns. You could solve it linearly, or just use matrix algebra.

Hmm, seems to me that still leaves me with variables instead of answers, I understand the current is constant (equal) but not how to calculate it exactly without more information.

You have 3 equations, three unknowns...... You can definately solve it.

The other two equations are:

R2/(R1+R2+R3) = 18/120

R1/(R1+R2+R3)=12/120


After:

you find all three R's via algebra (Matrix or linear)

add all three R's.....

I=V/R

P = IV or P = (V^2/R) or P = (I^2)R

 

[Quasmo]

Originally posted by: ChurchOfSubgenius

Originally posted by: Quasmo
If you use this you'll be irresistanceable.

Squeeze a small amount of product onto wet puff. Work it into a lather, massage over your body, and rinse.

Any man who knows what a "wet puff" is will be immediatly excused from the gender.

You obviously dont get the joke. Read carefully.