Question

I need help with part b of #1 and also #2

R_1 = 1.12 Ohm; R_2 = 2.24 Ohm; R_3 = 5.60 Ohm; R_4 = 3.36 Ohm How much current flows through each of the four resistors? I_1 = A I_2 = What is the total current delivered by the battery? What is the voltage across R_1? What are the voltages across each of the three parallel resistors? A I_3 = A I_4 = A

Answer 1

Q1.

part b:

if 3 resistances are connected in parallel,then if equivalent resistance is R,

1/R=(1/125)+(1/2500)+(1/4100)=8.6439*10^(-3)

==>R=115.688 ohms

Q2. R2,R3 and R4 are connected in parallel.

their net resistance let be R.

then 1/R=(1/R2)+(1/R3)+(1/R4)=(1/2.24)+(1/5.6)+(1/3.36)=0.92262

==>R=1.08387 ohms

R is in series with R1.

total resistance in the circuit=R+R1=1.08387+1.12=2.20387 ohms

current in the circuit=current through R=current through R1=V/2.20387=5.445 A

voltage across the parallel combination=current through R*value of R

=5.445*1.08387=5.90167 volts

then I2=5.90167/R2=2.63467 A

I3=5.90167/R3=1.05387 A

I4=5.90167/3.36=1.75645 A