Question

By considering the principles that the distribution of current in a circuit is the result of charge conservation, and the distribution of potential differences in a circuit is the result of energy conservation, derive:

(a) the combined resistance of two resistors R1 and R2 connected in series (4 marks)

(b) the combined resistance of two resistors R1 and R2 connected in parallel.

Answer 1

Explanations & Calculations

a)

• In the first circuit, potential difference (V) is distributed among the resistors while the same current flows through both of them (energy conserved: V= energy per unit charge)
• Therefore,

$\qquad\qquad \begin{aligned} \small V&= \small V_{R_1}+V_{R_2}\\ &= \small i_1R_1+i_1R_2\\ &= \small i(R_1+R_2) \end{aligned}$

• When the equivalent resistance is considered the same current should be flowing through it therefore,

$\qquad\qquad \begin{aligned} \small V&= \small iR_{eq} \end{aligned}$

• Equalling both the equations yield,

$\qquad\qquad \begin{aligned} \small iR_{eq}&= \small i((R_1+R_2)\\ \small R_{eq}&= \small R_1+R_2 \end{aligned}$

b)

• In a parallel connection, the same potential difference is sensed by both the resistors while the current is additive (charge conservation).
• Therefore,

$\qquad\qquad \begin{aligned} \small i_{total}&= \small i_1+i_2\\ &= \small \frac{V}{R_1}+\frac{V}{R_2}\\ \end{aligned}$

• When the equivalent resistance is considered for this circuit, the same total current should flow through it under the same potential difference.
• Therefore,

$\qquad\qquad \begin{aligned} \small i_{total}&= \small \frac{V}{R_{eq}}\\ \end{aligned}$

• As the 2 equations are combined it gives,

$\qquad\qquad \begin{aligned} \small \frac{V}{R_{eq}}&= \small \frac{V}{R_1}+\frac{V}{R_2}\\ \small \frac{1}{R_{eq}}&= \small \frac{1}{R_1}+\frac{1}{R_2}\\ \small R_{eq}&= \small \frac{R_1R_2}{R_1+R_2} \end{aligned}$