Electricity — Question 1

(a) 1 Ω and 10⁶ Ω are connected in parallel, their equivalent resistance will be:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{1}} + \dfrac{1}{10^6} = \dfrac{10^6 +1}{10^6}$

R_p = $\dfrac{10^6}{10^6 +1}$ ≈ $\dfrac{10^6}{10^6}$

R_p = 1 Ω

Therefore, the equivalent resistance is 1 Ω (approx).

(b) 1 Ω, 10³ Ω, and 10⁶ Ω are connected in parallel, their equivalent resistance will be:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{1}} + \dfrac{1}{10^3} + \dfrac{1}{10^6} = \dfrac{10^6 + 10^3 + 1}{10^6} = \dfrac{1001001}{1000000}$

R_p = $\dfrac{1000000}{1001001}$ = 0.999 Ω

Therefore, the equivalent resistance is 0.999 Ω.