Question

find the equivalent resistance of the following:​

Answer 1

Answer:

There are two sets of parallel resistances ( 1, 2, 3 ohm network and the 5 ohm, 6 ohm network ). We can understand they are in parallel as current must split to enter each resistor.

First we will calculate those two equivalent resistances.

For the first network, 1/R= 1/1 + 1/2 + 1/3 = 11/ 6

Therefore, R= 6/11 ohm

For the second network, 1/ R₂= 1/5 + 1/6= 11/ 30

Therefore, R₂= 30/ 11 ohm

Now 4 ohm, 6/ 11 ohm, 30/ 11 ohm and 7 ohm are in series as current flows without a break.

Total resistance= 4 + 6/11 + 30/ 11 + 7

                         = 157/ 11 ohm = 14. 2 ohm

Answer 2

Explanation:

Let resistance due to parallel arrangement of resistors of 1Ω , 2Ω and 3Ω be R'

$\frac{1}{R'} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R'} = \frac{6 + 3 + 2}{6} \\ \frac{1}{R'} = \frac{11}{6} \\ R' = \frac{6}{11}$

also, resistance due to parallel connection of resistors 5Ω & 6Ω is

$\frac{1}{R''} = \frac{1}{5} + \frac{1}{6} \\ \frac{1}{R''} = \frac{6 + 5}{30} \\ \frac{1}{R''} = \frac{11}{30} \\ R'' = \frac{30}{11}$

Total resistance in the circuit is

$R _{t} = R1 + R' + R'' + R4 \\ R _{t} = 4 + \frac{6}{11} + \frac{30}{11} + 7 \\ R_{t} = \frac{121 + 6 + 30}{11} \\R _{t} = \frac{157}{11} = 14.27Ω$