Question

There are four resistors of $12$ $ohm$ each. Which of the following values is/are possible by their combinations (series and / or parallel) ?

• A. $9$ $ohm$
• B. $16$ $ohm$
• C. $12$ $ohm$
• D. $30$ $ohm$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $9$ $ohm$
B $16$ $ohm$
C $12$ $ohm$
D $30$ $ohm$

1) In figure "A" we make a pair of 12$\Omega$ resistance in series and then connect them parallel to each other.
$R_{equivalent}=\frac{24\times 24}{24+24}=12\Omega$.
2)In figure "B" we connect first three 12$\Omega$ resistance in parallel to each other after that connect it with 12$\Omega$ resistance in series.
${\frac{1}{R}}_1=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{1}{4}$.
$R_{equivalent}=(4+12)\Omega =16\Omega$.
3)In figure "C" we connect first three 12$\Omega$ resistance in series and then combine it with single 12$\Omega$ resistance in parallel.
$R_{equivalent}=\frac{12\times 36}{12+36}=9\Omega$.
4)In figure "D" we have seen that we connect two 12$\Omega$ resistance in parallel and two 12$\Omega$ resistance in series and them make a combination to find it's equivalent.
$R_{equivalent}=\frac{12\times 12}{12+12}+24=30\Omega$.