Question

Combine three resistors 5$\sigma_{1}$, 4.5$\sigma_{2}$, and 3$\sigma_{1} + \sigma_{2}$ in such a way that the total resistance of this combination is maximum

• A. 12.5$\frac{2\sigma_{1}\sigma_{2}}{\sigma_{1} + \sigma_{2}}$
• B. 13.5 $I = 2x^{2} - 3t + 1$
• C. 14.5 $\overset{o}{A}$
• D. 16.5$9.4 \times 10^{18}$

Answer 1

Answer: (A)

12.5$\frac{2\sigma_{1}\sigma_{2}}{\sigma_{1} + \sigma_{2}}$

Answer 2

: For maximum equivalent or total resistance of the resistors must be combined in series.

$R_{eq} = R_{1} + R_{2} = R_{3}$

$= 5 + 4.5 + 3 = 12.5\Omega$