Question

The correct combination of three resistances $2.5 \times 10^{6}m^{2}V^{- 1}S^{- 1}$ and $2.5 \times 10^{4}m^{2}V^{- 1}S^{- 1}$ to get equivalent resistance $6.5 \times 10^{4}m^{2}V^{- 1}s^{- 1}$ is

• A. All three are combines in parallel
• B. All three are combine in series
• C. 1 $v_{d} \propto E$and 2 $v_{d} \propto E^{2}$ in parallel and 3$v_{d} \propto \sqrt{E}$ is in series to both
• D. 2 $v_{d} \propto \frac{1}{E}$ and 3 $\sigma_{1}$ are combined in parallel and 1$\sigma_{2}$ is in series to both

Answer 1

Answer: (D)

2 $v_{d} \propto \frac{1}{E}$ and 3 $\sigma_{1}$ are combined in parallel and 1$\sigma_{2}$ is in series to both

Answer 2

: According to option (4) $2\Omega$ and $3\Omega$ are combined in a parallel combination. Hence equivalent resistance.

}{\therefore R' = \frac{6}{5}}$$ $R'$is combined in series to $1\Omega$ $$\therefore R_{eq} = R' + 1 = \frac{6}{5} + 1 = \frac{6 + 5}{5} = \frac{11}{5}\Omega$$ Hence, the combination scheme in option (4) is correct.