Hi, I'm trying to figure out the current through each parallel branch of a series/parallel circuit. I can't draw the circuit so I'll describe it. We have a 100v battery, the current leaves the neg terminal of the battery travels along a single path hits R1 5ohms, then hits junction A where it splits into 3 parallel branches containing one resistor of 30 ohms each, R2, R3, R4 then combines at junction B travels along a single path where it hits R5 15 ohms, continues along a single path where it hits R6 20 ohms then back to the pos side of the battery.
Calculation of Rt: (R1) 5 ohms + Req(R2, R3, R4): 10 ohms ( 3 resistors of equal value 30/3 = 10) + (R5) 15 ohms + (R6) 20 ohms = 50 ohms. (5+10+15+20 = 50)
Calculation of It through mains: 100v/50ohms = 2 Amps
Now, the voltage drop of R1 is V=R*I or 5 ohms * 2 amps = 10 volts so R1 drops 10 volts. So, now at junction A we have 90v & still 2 amps right ? What is the current through each of the 30 ohm parallel branches ? Kirchoff's current law says whatever value of current enters a junction, the same value must leave the junction. So if 2 amps enters the junction at A, 2 amps must leave at junction B. But when I calculate the actual current through each branch using Ohms Law I get 3 amps each. 90v/30ohms of R2 = 3 amps across R2. Same calculation for the other 2 branches so 3 amps across R3 & 3 amps across R4. How can there be 2 amps entering, 2 amps leaving but 3 amps running along each branch in-between ? If I sum the 3 branches of 3 amps I'd get 9 amps.
What am I doing wrong ?
TIA
J