Show that if two resistors ${\mathbf{R}}_{\mathbf{1}}$and ${\mathbf{R}}_{\mathbf{2}}$are combined and one is much greater than the other $\left(R_1>>R_2\right)$: (a) Their series resistance is very nearly equal to the greater resistance ${\mathbf{R}}_{\mathbf{1}}$(b) Their parallel resistance is very nearly equal to smaller resistance${\mathbf{R}}_{\mathbf{2}}$ .

An electronic apparatus having a large capacitor to accommodate large amount of charge depending upon the requirement of the apparatus, even when the apparatus is switched off the capacitor present in the apparatus stores the charge and when touched directly presents a shock hazard. In order to avoid the shock, a bleeder resistor is placed across such a capacitor to take out the charge which is stored in the capacitor when the apparatus is switched off.

Consider two resistors linked in series or parallel in the limit , ${\mathrm{R}}_1>>{\mathrm{R}}_2$say and${\mathrm{R}}_1=1000\mathrm{\Omega }\mathrm{and}{\mathrm{R}}_2=1\mathrm{\Omega }$

(a)

Because the overall resistance of a series of resistors is equal to the sum of their individual resistances, therefore,

$$\begin{gathered} R=R_1+R_2 \\ =1000\Omega +1\Omega \\ =1001\Omega \end{gathered}$$

This is very close to${\mathrm{R}}_1$

The difference in relative terms is

$$\begin{gathered} \frac{\left|R_1-R\right|}{R_1}=\frac{1}{1000} \\ =0.1\% \end{gathered}$$

This is understandable because the same current goes through the circuit and is impacted by the overall resistance, which is dominated by the largest resistance.

(b)

The total resistance is equal to the number of resistors connected in parallel.

$$\begin{gathered} \frac{1}{\mathrm{R}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2} \\ \frac{1}{\mathrm{R}}=\frac{1}{1000\mathrm{\Omega }}+\frac{1}{1\mathrm{\Omega }} \\ \mathrm{R}=0.999\mathrm{\Omega } \end{gathered}$$

here due to small number $\frac{1}{{\mathrm{R}}_1}$is neglected

$$\begin{gathered} \frac{\left|R_2-R\right|}{R_2}=\left(\frac{9.99\times {10}^{-4}}{1}\right)\times 100\% \\ =0.0999\% \end{gathered}$$

This is close to$R_2$ , with the relative difference

Because current follows the path of least resistance, the branch with the highest resistance will essentially be omitted from the circuit, and the majority of the current will pass via the branch with the lowest resistance.