# Practical Electronics/Resistors/Parallel Proof

This page presents a proof that the resistance of a parallel network, Rtot of n resistors is the reciprocal of the sum of the reciprocals of the individual resistors:

First, consider that a current into (and out of) the whole network, itot, must be equal to the sum of the currents in each resistor (Kirchhoff's Current Law).

${\displaystyle i_{tot}=\sum \limits _{k=1}^{n}{i_{k}},}$

where ik is the current in the kth resistor in the network. Now, from Ohm's Law, we can write:

${\displaystyle {{v_{tot}} \over {R_{tot}}}=\sum \limits _{k=1}^{n}{{v_{k}} \over {R_{k}}},}$

where:

• vtot is the voltage across the entire network
• Rtot is the resistance of the entire network
• vk is the voltage across the kth resistor in the network
• Rk is the resistance of the kth resistor in the network

Since we know that the voltage across each resistor is the same (a trivial case of Kirchhoff's Voltage Law), we can say

${\displaystyle {{v} \over {R_{tot}}}=\sum \limits _{k=1}^{n}{{v} \over {R_{k}}},}$

where v is the voltage across the network. Dividing through by v gives us our final answer:

${\displaystyle {{1} \over {R_{tot}}}=\sum \limits _{k=1}^{n}{{1} \over {R_{k}}}.}$

This is also sometimes equivalently written as:

${\displaystyle R_{tot}={1 \over {\sum \limits _{k=1}^{n}{1 \over {R_{k}}}}}}$

This is a vital result and should be remembered. Notice that the proof for series networks is essentially the same, but uses KCL first and KVL second.

Other interesting points are that this result describes behaviour of series capacitive networks and parallel inductive networks.