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Two resistors $R_{1}$ and $R_{2}$ are connected in series. Their combined effect is $16 Ω$ . If the difference between the resistance of these resistors is $8 Ω$ . Which of the following pairs is the correct pair of resistance?

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R_{1} = 8 ohms, R_{2} = 8 ohms

R_{1} = 12 ohms, R_{2} = 4 ohms

R_{1} = 10 ohms, R_{2} = 6 ohms

R_{1} = 15 ohms, R_{2} = 1 ohms

answer is B.

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Detailed Solution

The correct option is

2

.
Given:

R_{1} + R_{2} = 16 Ω

and

R_{1} - R_{2} = 8 Ω

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}

It is given that

R_{1} + R_{2} = 16 Ω

and also

R_{1} - R_{2} = 8 Ω

.
On adding both the equations, we get

R_{1} = 12 Ω

Substituting the value of

R_{1}

in any of the equation, we get the value of

R_{2}

R_{2} = 4 Ω

Hence, the correct pair is

R_{1} = 12 ohms, R_{2} = 4 ohms

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