Using Kirchoff rule to find Voltage drop and current, in a complex circuit

[immyrichie]

Homework Statement

Measure the Voltage drop and current in all element of the circuit, given the figure which has the following Resistance and its Ideal emfs;

R1=20ohms, R2=70ohms, R3=55ohms, R4=45ohms, R5=30ohms, R6=20ohms, R7=70ohms, E1=15v, E2=5v, E3=5v.

Homework Equations

The Attempt at a Solution

please I need help as this is my first time coming across the Kirchoff law on a problem, i tried writing the follow after I read some books but it seems confusing if am correct thus i can't proceed further

I1+13=I2
I2+15=14+16


E3=I6R6+I5
E2=I3R3+I2R2+I4R4

 

[tiny-tim]

welcome to pf!

hi immyrichie! ! welcome to pf!

(try using the X₂ icon just above the Reply box )

I1+13=I2
I2+15=14+16

fine

E3=I6R6+I5
E2=I3R3+I2R2+I4R4

noooo …

the first one should be just E₃ = I₆R₆ …

you need the voltage drop across each component: since there's no component in the I₅ section, it doesn't appear (and you're only allowed IR combinations anyway, not an I on its own )

in the second one, you've left out I₃R₇

(and of course there's two more KVL loops to do)

 

[gneill]

It might be worthwhile to consider given that the voltage sources are all ideal (no internal resistance), that E3 fixes the voltage across R4 and across R6. Thus via Ohm's law, the currents through both resistors are also fixed. This means that R4 and R6 are "solved" for their voltages and currents.

It also means that R4 and R6 play no further role in the circuits analysis and can be ignored (you can actually remove them from the circuit without affecting anything else). This is so because again, E3 is completely determining the voltage across the R4/R6/E3 combination, and thus only E3 will appear in any KVL equation that includes them as a "component". This leaves you with just two KVL loops to handle.

 

[immyrichie]

thanks so much for reply, since I'm now left with two loop I tried this out for the loops;

E2=I3R3+I2R2+I4R4+I3R7;
E1= I1R1-I2R2-I4R4-I3R5.

how well is this?

 

[SammyS]

thanks so much for reply, since I'm now left with two loop I tried this out for the loops;

E2=I3R3+I2R2+I4R4+I3R7;
E1= I1R1-I2R2-I4R4-I3R5.

how well is this?

The first one looks OK.

The second one has errors.

The current through R₅ is I₁.
You have the wrong signs on I₂R₂ & I₄R₄ & what should be I₁R₅.​

Don't forget: As gneill pointed out, I₄R₄ = E₃ = 5V.

 

[tiny-tim]

hi immyrichie!

(just got up :zzz: …)

just to add to what SammyS says …

you can take any loop, so the simplest ones would be the ones with the fewest Rs …

ie the loop containing E₁ R₂ E₃ and R₅, and the loop containing E₂ R₃ R₂ E₃ and R₇

 

[immyrichie]

hah, thanks so much guys. after these equation, what's next?

 

[tiny-tim]

put all the E and R numbers in, and solve for all the Is

then the voltage drop across each component is IR