If two resistors are connected in series, the total resistance is $45 Ω$ and if the same resistors are connected in parallel the total resistance becomes $10 Ω$ . What is the value of individual resistors ?

15 Ω, 30 Ω

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15 Ω, 10 Ω

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15 Ω, 20 Ω

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45 Ω, 30 Ω

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Solution

The correct option is A $15 Ω, 30 Ω$
The total resistance of resistors in series is equal to the sum of their individual resistances. That is, $R_{t o t a l} = R_{1} + R_{2} + R_{3}$ .
The total resistance will always be less than the value of the smallest resistance That is, $\frac{1}{R_{t o t a l}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} t h a t i s, R = {R_{t o t a l}}^{- 1}$ .
In this case, the total resistance of the resistors connected in series is given as $R_{1} + R_{2} = 45$ - eqn 1
The total resistance of the resistors connected in parallel is given as
$\frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{10} t h a t i s, \frac{1}{R_{1}} + \frac{1}{R_{2}} = 0.1$ - eqn 2
Putting the value of $R_{1} + R_{2}$ in eqn 2, we get, $R_{1} R_{2} = 10 \times 45 = 450 o h m s$ .
Now, From the identity, ${{{{(R}_{1} {- R}_{2})}^{2} = (R}_{1} {+ R}_{2})}^{2} - 4 R_{1} R_{2}, w e g e t, R_{1} {- R}_{2} = 15 o h m s$ - eqn 3.
Adding eqn 2 and eqn 3,
${2 R}_{1} = 60 o h m s T h a t i s, R_{1} = 30 o h m s$ .
Substituting this value in eqn 3, we get, $R_{2} = (30 - 15) = 15 o h m s$ .
Hence, the values of the resistors are 30 ohms and 15 ohms.

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If two resistors are connected in series, the total resistance is 45Ω and if the same resistors are connected in parallel the total resistance becomes 10Ω. What is the value of individual resistors ?

If two resistors are connected in series, the total resistance is $45 Ω$ and if the same resistors are connected in parallel the total resistance becomes $10 Ω$ . What is the value of individual resistors ?