Question

Five equal resistances of $I = 2x^{2} - 3t + 1$ are connected between A and B as shown in figure. The resultant resistance is

• A. 10 $\overset{o}{A}$
• B. 5$9.4 \times 10^{18}$
• C. 15$\left( e = 1.6 \times 10^{- 19}C \right)$
• D. 6$4\Omega$

Answer 1

Answer: (B)

5$9.4 \times 10^{18}$

Answer 2

: According to the given circuit $10\Omega$ and $10\Omega$ resistances are connected in series.

$\therefore R' = 10 + 10 = 20\Omega$

Again $10\Omega$and $10\Omega$ resistances are connected in series

$\therefore R'' = 10 + 10 = 20\Omega$

$R',R''$ and $10\Omega$ all connected in parallel than

$\therefore\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R''} + \frac{1}{10} = \frac{1}{20} + \frac{1}{20} + \frac{1}{10} = \frac{1 + 1 + 2}{20}$

$= \frac{4}{20} = \frac{1}{5}$

$R_{eq} = 5\Omega$