# Kirchhoffs Rule?

0

The circuit to the right consists of a battery (V0 = 42.5 V) and five resistors (R1 = 111 Ω, R2 = 682 Ω, R3 = 263 Ω, R4 = 634 Ω, and R5 = 265 Ω). Find the current passing through each of the specified points.

• E=42.5V, R1=111Ω,R2=682Ω,R3=263Ω,R4=634Ω,R5=265Ω

R23= (682*263)/(682+263)= 189.805Ω

R45= (634*265)/(634+265)= 186.885Ω

R2345= (189.805*186.885)/(189.805+186.885)= 94.167Ω

R12345= 111+94.166= 205.167Ω (Equivalent resistance)

I= 42.5/205.167= 0.207148 = 207.148mA

E2345= 42.5*94.167/205.167= 19.506V

Ia=E2345/R2 = 19.506/682= 0.028601= 28.6[mA]

Ic=E2345/R45= 19.506/186.885= 0.104374 = 104.4[mA]

If=I-Ia = 0.207148-0.028601= 0.178547= 178.5[mA]

• First determine the resistance seen by the voltage source:

Req = 111Ω+ 1/(1/682+1/263+1/634+1/265) = 111+94.167 = 205.167Ω

The current leaving the source must be V/Req = 42.5/205.167 = 0.2071A = 207.1ma

The remaining 4 resistors are in parallel which means they all have the same voltage across them.

What is that voltage? It is 42.5V – 0.2071A*111Ω = 19.51V

Knowing the voltage across each resistors allows us to calculate the current in each

iR1 = 207.1ma

iR2 = 19.51V/682Ω = 28.6ma = iA

iR3 = 19.51V/263Ω = 74.2ma

iR4 = 19.51V/634Ω = 30.8ma

iR5 = 19.51V/265Ω = 73.6ma

iA = iR2

iC = iR4+iR5 = 104.4ma

iF = iR1-iR2 = 178.5ma = iR3+iR4+iR5 =178.6ma

So we have a little round off error there.

Hope this helps

Also Check This  theres a a mexican song?