
The equivalent resistance of series combination of four equal resistors is S. If they are joined in parallel, the total resistance is P. the relation between S and P is given by S = nP, then the minimum possible value of n is?
A. 12
B. 14
C. 16
D. 10
Answer
576.3k+ views
Hint: This question deals with relation between 4 resistors when connected in series and later in parallel. Two resistors are said to be in parallel when the nodes of both ends are the same. If only one node is the same the circuit is said to be in series combination. Here, the unknown variable is the resistance of the resistor. So, in order to solve this question, we must assume that all the resistors have equal resistance and it is equal to R. So, we will first find the equivalent resistance when the series is connected in series S then in parallel P. We will then substitute these values in the given equation to calculate our answer.
Formula Used:
S = nP
\[S={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+{{R}_{4}}\]
\[\dfrac{1}{P}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}+\dfrac{1}{{{R}_{4}}}\]
Complete step-by-step answer:
As we know, if the resistors are connected in series the equivalent resistance is given by,
\[S={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+{{R}_{4}}\]
Now, assume that all the resistors have same resistance and equal to R
So, the above equation becomes
\[S=R+R+R+R=4R\]
\[R=\dfrac{S}{4}\]……. (1)
Similarly, if the resistors are in parallel the equivalent resistance is given by,
\[\dfrac{1}{P}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}+\dfrac{1}{{{R}_{4}}}\]
\[\dfrac{1}{P}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}\]
\[\therefore P=\dfrac{R}{4}\]
\[R=4P\]………… (2)
From (1) and (2)
\[\dfrac{S}{4}=4P\]
Solving the above equation,
We get,
\[\therefore S=16P\]
Therefore, the correct answer is option C.
Note: There is no alternative way to solve this specific question. Here the unknown variable is the resistance of the resistor. So, in order to solve this question we must assume that all the resistors have equal resistance and it is equal to R.
Formula Used:
S = nP
\[S={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+{{R}_{4}}\]
\[\dfrac{1}{P}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}+\dfrac{1}{{{R}_{4}}}\]
Complete step-by-step answer:
As we know, if the resistors are connected in series the equivalent resistance is given by,
\[S={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+{{R}_{4}}\]
Now, assume that all the resistors have same resistance and equal to R
So, the above equation becomes
\[S=R+R+R+R=4R\]
\[R=\dfrac{S}{4}\]……. (1)
Similarly, if the resistors are in parallel the equivalent resistance is given by,
\[\dfrac{1}{P}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}+\dfrac{1}{{{R}_{4}}}\]
\[\dfrac{1}{P}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}\]
\[\therefore P=\dfrac{R}{4}\]
\[R=4P\]………… (2)
From (1) and (2)
\[\dfrac{S}{4}=4P\]
Solving the above equation,
We get,
\[\therefore S=16P\]
Therefore, the correct answer is option C.
Note: There is no alternative way to solve this specific question. Here the unknown variable is the resistance of the resistor. So, in order to solve this question we must assume that all the resistors have equal resistance and it is equal to R.
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