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Question 101 of 102

Q.Two resistors when connected in series and parallel, their equivalent resistances are 15 Ω\Omega and 5615 Ω\dfrac{56}{15}\ \Omega respectively. Find the values of the resistances.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2026Subjective· 3mImportance★★★★★
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Setting up the sum and product of the two resistances from the series and parallel equivalents and solving the resulting quadratic gives R1=8 ΩR_1=8\,\Omega and R2=7 ΩR_2=7\,\Omega.

Working

Series: R1+R2=15R_1+R_2 = 15 ... (1)

Parallel: R1R2R1+R2=5615\dfrac{R_1R_2}{R_1+R_2} = \dfrac{56}{15}. Using (1):

R1R215=5615 ⇒ R1R2=56...(2)\dfrac{R_1R_2}{15} = \dfrac{56}{15} \ \Rightarrow\ R_1R_2 = 56 \quad \text{...(2)}

Solving. R1,R2R_1,R_2 are roots of x2−(R1+R2)x+R1R2=0x^2-(R_1+R_2)x+R_1R_2=0:

x2−15x+56=0x^2 - 15x + 56 = 0

Using the quadratic formula: …

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