Question

12. Ry=10 Ohm, R2 = 20 Ohm, R3 = 30 Ohm. Calculate the effective resistance whenthey are connected in (i) series (ii) parallel. How can we connect these resistancesto produce a resultant resistance of 22 Ohm? Justify the answer.​

Answer 1

Answer:

1. Series

Req. = R1+ R2+R3

= 10+20+30

=60ohms

2 Parallel

$\frac{1}{r \: eq} = \frac{1}{r1} + \frac{1}{r2} + \frac{1}{r3} \\ \\ = \frac{1}{10} + \frac{1}{20} + \frac{1}{30} \\ \\ = \frac{6}{60} + \frac{3}{60} + \frac{2}{60} \\ \\ = \frac{6 + 3 + 2}{60} \\ \\ = \frac{11}{60} \\ \\ \frac{1}{r \: eq} = \frac{11}{60} \\ \\ r \: eq = \frac{60}{11} = 5.36ohm$

3. For an equivalent resistance of 22 ohm we should connect R2(20 ohm) and R3(30ohm) in parallel then this combination is in series with R1(10ohm) as shown in the figure drawn.

Therefore the equivalent resistance of R1 and R2is

12ohm. Now this combination is series with R1. Therefore total resistance 22ohm.

I hope you are clear if not just comment me and mark me as the brainliest.