Question
You are given several identical resistances each of value $$R = 10\,\Omega $$ and each capable of carrying a maximum current of $$1\,A.$$ It is required to make a suitable combination of these resistances of $$5\,\Omega $$ which can carry a current of $$4\,A.$$ The minimum number of resistances of the type $$R$$ that will be required for this job is
A.
4
B.
10
C.
8
D.
20
Answer :
8
Solution :
To carry a current of $$4\,A$$ we need four path, each carrying a current of $$1\,A.$$ Let $$r$$ be the resistance of each path. These resistances are connected in parallel. So, their equivalent resistance.
$$\eqalign{ & \frac{1}{{{r_p}}} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} \cr & {\text{or}}\,\,{r_p} = \frac{r}{4}. \cr & {\text{But,}}\,\,{r_p} = \frac{r}{4} = 5\,\,\left( {{\text{given}}} \right) \cr & \therefore r = 20\,\Omega \cr} $$
For this purpose two resistances should be connected is series. There are four such combinations in parallel. Hence, the total number of resistances $$ = 4 \times 2 = 8.$$
To carry a current of $$4\,A$$ we need four path, each carrying a current of $$1\,A.$$ Let $$r$$ be the resistance of each path. These resistances are connected in parallel. So, their equivalent resistance.
$$\eqalign{ & \frac{1}{{{r_p}}} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} \cr & {\text{or}}\,\,{r_p} = \frac{r}{4}. \cr & {\text{But,}}\,\,{r_p} = \frac{r}{4} = 5\,\,\left( {{\text{given}}} \right) \cr & \therefore r = 20\,\Omega \cr} $$
For this purpose two resistances should be connected is series. There are four such combinations in parallel. Hence, the total number of resistances $$ = 4 \times 2 = 8.$$
