# Question #8e836

Jan 31, 2018

$\text{See explanation}$

#### Explanation:

$\text{The sum of the current in both parallel branches is equal}$
$\text{to the ingoing and outgoing current I :}$
$I = \frac{V}{R} = I 1 + I 2 = \frac{V}{R 1} + \frac{V}{R 2} = V \left(\frac{1}{R 1} + \frac{1}{R 2}\right)$
$\implies \frac{1}{R} = \frac{1}{R 1} + \frac{1}{R 2}$

Jan 31, 2018

The proof is only true for a parallel circuit:

#### Explanation:

Please observe that in the above parallel circuit the voltage across the two resistors is the same value, V.

The current equation for the node where the voltage source and the two resistors are connected is:

$I = {I}_{1} + {I}_{2} \text{ [1]}$

Use Ohm's Law to write the equations for the currents ${I}_{1}$ and ${I}_{2}$ in terms of the voltage, V, and its respective resistance for both currents are:

${I}_{1} = \frac{V}{R} _ 1 \text{ [2]}$

${I}_{2} = \frac{V}{R} _ 2 \text{ [3]}$

Substitute equations [2] and [3] into equation [1] and designate it [1.1]:

$I = \frac{V}{R} _ 1 + \frac{V}{R} _ 2 \text{ [1.1]}$

Use Ohm's Law to write equation for the current, I, in terms of the voltage V and an equivalent resistance $R$:

$I = \frac{V}{R} \text{ [4]}$

Substitute equation [4] into equation [1.1] and designate it [1.2]:

$\frac{V}{R} = \frac{V}{R} _ 1 + \frac{V}{R} _ 2 \text{ [1.2]}$

Please observe that V is a common factor to both sides and can be eliminated by division:

$\frac{1}{R} = \frac{1}{R} _ 1 + \frac{1}{R} _ 2 \text{ [1.3]}$ Q.E.D.

NOTE: the above equation is extendable to any amount of resistors in parallel:

$\frac{1}{R} = \frac{1}{R} _ 1 + \frac{1}{R} _ 2 + \frac{1}{R} _ 3 + \ldots \text{ [1.4]}$

Feb 26, 2018

Use the geometric definition of resistance of a resistor

#### Explanation:

Resistance = Resistivity*Length / Area
...
When you add two or three (or any number) of resistors in parallel, you are geometrically adding cross sectional area directly (which is in the denominator of the resistance formula).
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I published an instructional video on the topic that may be helpful: