Find total resistance for the following circuit

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find the total resistance from AtoB, i'm not sure if i add everything even the 6amps?

It is like this as we can see the resistances 5,6 and 12 are in parallel. So, to find their equivalent resistance do like this

1/R1 = 1/5 + 1/6 + 1/12

After solving this, we get

R1 = 20/9

Again, 4 and 8 are in parallel. So their equivalent will be

R2 = (4 x 8)/(4 + 8)
= 8/3
R3 = 4


To get R we add all R1, R2 and R3

R = R1 + R2 + R3
= 20/9 + 8/3 + 4
=80/9 Ans.

The 4 ohm resistor on the left is in series with two groups of parallel resistors.

The first group of three resistors calculates to a total resistance of 2.222 ohms

The second group; of two resistors calculates to 2.666 ohms

The computed values continue the series resistance, additively:

4 ohms + 2.222 ohms + 2.666 ohms = 8.888 ohms

The formula for RT of TWO resistors in parallel is (R1 * R2) / (R1 + R2) which equals 2.666 ohms here

RT being the total equivalent resistance -- these two resistors could be replaced with ONE 2.666 ohm resistor without changing the circuit electrically

The formula for more than two resistors is a bit more complex, but I used the calculator here:

http://www.csgnetwork.com/parallelresist…

AND that calculator give the EXACT formula for THREE resistors in parallel.


You ADD resistances in series to find the Total Resistance.

When resistors are in parallel, the total resistance is some value smaller than the resistance of the smallest resistor involved.

5,6,12ohms are parallel .4&8ohms are also parallel .these two combinations are series with 4ohms
1/5+1/6+1/12=27/60(sum of reciprocal is resistances since they are parallel)
1/4+1/8=3/8
27/60+3/8+4=193/40=4.825ohms(sum of resistances when they are in series )