Question

A circuit consists of a battery and two 22-Ω resistors in series. The current through the circuit is 0.55 A. A third resistor is then added to the circuit in parallel with the first two resistors. The current through the new branch of the circuit is 0.35 A. What is the resistance of the third resistor?

Answer 1

Answer:

C=E/R

In the 1st case 0.55=E/44

E=0.55×44

Let r be the 3rd resistance. This r is in parallel to 44.

Then effective resistance R in second case is

=1/(1/44+1/r) =44r/(44+r)

0.35=0.55×44/{44r/(44+r)}

0.35r=0.55r+44×0.55

- 0.20r=44×0.55

r =44×0.55/0.20= -121

you are getting negative sign because after adding 3rd resistance in parallel, the effective resistance will decrease and the current will increase and become more than o.55. If the 3rd resistance is put in series with 44 ohm then current will be less than 0.55