# Question 917b3

Dec 3, 2017

Parallel Circuits: For conservative force fields, one can define a scalar potential. As a charge moves from one point to another, the change in electric potential is independent of the path but depends only on the end points.

In a parallel connection the end points of the different branches are the same. So charges going through different parallel path will all experience the same potential difference.

The current through each brach depends on the resistance through that path. Paths with lower resistance carry higher current (because charges can move relatively easily through them) and viceversa.

Series Circuits: In a series circuit the end points of the different resistors are different. If we consider a path from point A to point E with intermediate points B, C and D.

Then the potential difference between A and E is -
\DeltaV_{AE} = \int_{AE}dV = V_E - V_A;# ...... (1)

$\setminus {\int}_{A E} \mathrm{dV} = \setminus {\int}_{A B} \mathrm{dV} + \setminus {\int}_{B C} \mathrm{dV} + \setminus {\int}_{C D} \mathrm{dV} + \setminus {\int}_{D E} \mathrm{dV}$

$\setminus \Delta {V}_{A E} = \setminus \Delta {V}_{A B} + \setminus \Delta {V}_{B C} + \setminus \Delta {V}_{C D} + \setminus \Delta {V}_{D} E$

$\setminus q \quad \setminus q \quad \setminus \quad = \left(\cancel{{V}_{B}} - {V}_{A}\right) + \left(\cancel{{V}_{C}} - \cancel{{V}_{B}}\right) + \left(\cancel{{V}_{D}} - \cancel{{V}_{C}}\right)$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad + \left({V}_{E} - \cancel{{V}_{D}}\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad = {V}_{E} - {V}_{A} = \setminus {\Delta}_{A E}$

The current through all the resistors are same, because of charge conservation. Charge accumulation or depletion cannot happen anywhere inside the circuit. What goes-in must come out. So if a certain number of charge carrier gets through point A, the same number of charges must pop out through point E.