Maths - Resistors in series and parallel: Intermediate

[D Skelton]

Maths - Resistors in series and parallel: Intermediate

 

[Baker1988]

i dont understand what the two lines with 13 volts is

 

[Happy Jack]

i dont understand what the two lines with 13 volts is

That's the voltage measured or volt drop across the resistor.

 

[Baker1988]

That's the voltage measured or volt drop across the resistor.

so he has put the volt drop on the diagream

 

[Robson689]

Voltage in a parallel circuit is constant right, but those two resistors at the top are in series where voltage is not constant so it would be different across the two resistors. That's how I read it anyway...

Since voltage is constant I would work out first the voltage at C, and then work out the resistance at A which would enable you to work out the current at B and calculate the total resistance.

V = IR so I = V/R.

13 / 3 = 4.33A and because it is series at the top, it would be constant across both resistors so you can work out the voltage at the top left by using V = IR which would give you 4.33 x 33 = 142.89v.

Since voltage is constant, would that mean that the voltage at C would be 142.89v?

If it is, the resistance at A would be 6.8Ω (R = V/I = 142.89/21).

The current at B would be worked out by working out the current at the top and adding it to 21 right? So because I = V/R you would do 142.89/33 = 4.33A and that plus the 21A would give you a total of 25.33A at B.

Since you work out resistance in parallel as 1/R(total) = 1/R1 + 1/R2 you would do:
1/33 = 0.03
1/6.8 = 0.15

Add those together and divide by 1 gives you 5.56Ω for the total resistance which would make sense because the total resistance in parallel is less than the value of the smallest resistor.


Sorry for the mess, I just typed it up as I figured it out.

Total Resistance (Rt) = 5.56Ω
Resistance at A = 6.8Ω
Current at B = 25.33A
Voltage at C = 142.89V

 

[Happy Jack]

so he has put the volt drop on the diagream

That's it. You need both of those pieces of information to solve for the current through that series leg of the circuit.

 

[Happy Jack]

Voltage in a parallel circuit is constant right, but those two resistors at the top are in series where voltage is not constant so it would be different across the two resistors. That's how I read it anyway...

You solve Ohms Law with I as the subject, so I = V / R, or 13 / 3, which gives you the current through both of the resistors in series there. You can then solve for V as you have both I and R, so the voltage across the 30R resistor would be V = (13 / 3) x 30. The rest should fall into place.

What I like about these sort of questions is that the actual maths is childs play. It is the ability to juggle Ohms Law, remember and apply the rules for both voltage and current in series / parallel, KVL and KCL, and the process of reducing a variety of the resistances in parallel to the series equivalent, much like the Thevenin method, that make it challenging.

 

[D Skelton]

Total Resistance (Rt) = 5.56Ω
Resistance at A = 6.8Ω
Current at B = 25.33A
Voltage at C = 142.89V

Absolutely correct! Your working out is good too.

Why not try your hand at the advanced one

 

[Baker1988]

Absolutely correct! Your working out is good too.

Why not try your hand at the advanced one

i just worked this one out before i scrolled down and saw answers (so i didnt cheat)and that and i got this

Resistance total (Rt) 5.54 ohms
resistance at A = 6.8 ohms
current at B = 25.72 A
voltage at C = 142.5 V