Question

There are four resistors of 12 ohm each.
Which of the following values is/are possible
by their combinations (series and/or parallel)?
[IJSO/Stage-1/2008]
(A) 9 ohm
(B) 16 ohm
(C) 12 ohm
(D) 30 ohm​

Answer 1

Answer:

Only 16 & 30

Explanation:

• If we Join 2 Resitor In Series And thier Equivalent Resistance Parllel To It. Will come out to be 30
• $\frac{1}{r1} + \frac{1}{r2} + r3 + r4 = 30$
• If we join three in Parllel and one in series with their equivalent
• $\frac{1}{r1} + \frac{1}{r2} + \frac{1}{r3} + r4 = 16$

Hope This Will Help..

Answer 2

Answer:

only 16 and 30 are possible

Explanation:

if we take 2 parallel and 2 in series, then

1/ R equivalent = 1/R1 +1/R2

= 1/12 +1/12

= 2/12 = 1/6

then R equivalent= 6(parallel)

,and there are two in series

so , 12+12 = 24 (series)

therefore , total will be

series+parallel

24+6

= 30

same like that

take 3 parallel combinations and 1 in series

after simplying that we get our answer as 16

( pls solve it by urself , according to above )

therefore, only 16 and 30 will be the answer