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Two resistances of the same length and the same materials are first connected in parallel and then in series. What is the ratio of the equivalent resistance in both cases?

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1 : 3

1 : 4

1 : 2

1 : 1.5

answer is B.

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Detailed Solution

The ratio of the equivalent resistance in both cases is given by

1 : 4

.
Let the equivalent resistances of the wires if connected in series be

R_{S}

, the equivalent resistances of the wires if connected in parallel be

R_{P}

and let the resistances be

R

.
It is known that “in a series combination, the equivalent resistance is equal to the sum of the individual resistances.” Mathematically,

R_{S} = R_{1} + R_{2} + . . . + R_{n}

Where

R_{S}

is the series equivalent resistances and

R_{1}, R_{2}, . . ., R_{n}

are the individual resistances.
For two resistances, the equivalent resistance in the series combination is given by

R_{S} = R_{1} + R_{2}

On substituting the given resistances, we get the series equivalent resistance as

{\Rightarrow R}_{S} = R + R

{\Rightarrow R}_{S} = 2 R

Also, “in a parallel combination, the reciprocal of the equivalent resistance is equal to the sum of the reciprocal of the individual resistances.” Mathematically,

\frac{1}{R_{P}} = \frac{1}{R_{!}} + \frac{1}{R_{2}} + . . . + \frac{1}{R_{n}}

Where

R_{p}

is the parallel equivalent resistances and

R_{1}, R_{2}, . . ., . R_{n}

are the individual resistances.
For two resistances, the equivalent resistance in the parallel combination is given by

\frac{1}{R_{P}} = \frac{1}{R_{!}} + \frac{1}{R_{2}}

On substituting the given resistances, we get the parallel equivalent resistance as

\Rightarrow \frac{1}{R_{P}} = \frac{1}{R} + \frac{1}{R}

\Rightarrow \frac{1}{R_{P}} = \frac{2}{R}

\Rightarrow R_{P} = \frac{R}{2}

Hence, the ratio of the equivalent resistance in both cases is given by

\Rightarrow \frac{R_{P}}{R_{S}} = \frac{(\frac{R}{2})}{2 R}

\Rightarrow \frac{R_{P}}{R_{S}} = \frac{1}{4}

Therefore, the ratio of equivalent resistance is

1 : 4

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