Remember the junction rule current IN current OUT In the se

Remember the junction rule: current IN = current OUT In the section of circuit below R_1 = 5.500 ohm, R_2 = 10, 300 ohm, and the voltage difference v_a - v_b = 4.500 V. If the total current in the circuit i = 0.340 A. find the value of R_3 correct, computer gets:

Solution

For solving such problems we need to keep following two points into consideration:

1.) For resitors R1, R2 and so on connected in series, the net resistance is given as Rnet = R1 + R2 and so on

2.) For the resistors to be connected in parallel, the net resistance is given as 1 / Rnet = 1/R1 + 1/R2 and so on.

3.) Current in the components connected in series is same while potential differnce across components connected in parallel are same.

We will make use of the above to solve the given problem as follows:

As the R2 and R3 are in parallel, we can write for combined resistance as:

1 / R₂₃ = 1/ R2 + 1/R3

or, R₂₃ = R2 x R3 / R2 + R3 = 10.3R3 / (10.3 + R3)

Also this R₂₃ is in series with R1, hence we have Rnet = R1 + 10.3R3 / (10.3 + R3)

Now we know that we have a potential difference across a and b as V = 4.5

and we have the relation for current and voltage as: I = V/R

So we can write: 0.54 = 4.5 / Rnet

or, Rnet = 8.333

or, R1 + 10.3R3 / (10.3 + R3) = 8.333

or, 10.3R3 / (10.3 + R3) = 2.8333

or,10.3R3 = 29.1833 + 2.8333R3

or, 7.1667R3 = 29.1833

or, R3 = 4.0721

Therefore the required value of R3 is 4.0721 ohms