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- https://phys.libretexts.org/Courses/Coalinga_College/Physical_Science_for_Educators_(CID%3A_PHYS_14)/11%3A_Electricity/11.04%3A_Electric_Current_and_Resistance/11.4.04%3A_Ohm's_LawThe units of resistance can be determined using the units of the other terms in the equation, namely that the potential difference is in volts (J/C) and current in amperes (C/s): Since the current is ...The units of resistance can be determined using the units of the other terms in the equation, namely that the potential difference is in volts (J/C) and current in amperes (C/s): Since the current is directly proportional to the potential difference and inversely proportional to the resistance, you can increase the current in a circuit by increasing the potential or by decreasing the resistance.
- https://phys.libretexts.org/Courses/Coalinga_College/Physical_Science_for_Educators_(CID%3A_PHYS_14)/11%3A_Electricity/11.04%3A_Electric_Current_and_Resistance/11.4.03%3A_Ohms_Law-_Resistance_and_Simple_CircuitsFigure \(\PageIndex{1}\): A simple electric circuit in which a closed path for current to flow is supplied by conductors (usually metal wires) connecting a load to the terminals of a battery, represen...Figure \(\PageIndex{1}\): A simple electric circuit in which a closed path for current to flow is supplied by conductors (usually metal wires) connecting a load to the terminals of a battery, represented by the red parallel lines. In a simple circuit (one with a single simple resistor), the voltage supplied by the source equals the voltage drop across the resistor, since \(\mathrm{PE}=q \Delta V\), and the same \(q\) flows through each.
- https://phys.libretexts.org/Bookshelves/Thermodynamics_and_Statistical_Mechanics/Essential_Graduate_Physics_-_Statistical_Mechanics_(Likharev)/06%3A_Elements_of_Kinetics/6.02%3A_The_Ohm_law_and_the_Drude_formulaNow separating the vector \(\mathbf{v}\) outside the parentheses into the component \(v \cos\theta\) directed along the vector \(\pmb{\mathscr{E}}\), and two perpendicular components, \(v \sin\theta \...Now separating the vector \(\mathbf{v}\) outside the parentheses into the component \(v \cos\theta\) directed along the vector \(\pmb{\mathscr{E}}\), and two perpendicular components, \(v \sin\theta \cos\varphi\) and \(v\sin\theta \sin\varphi \), we see that the integrals of the last two components over the angle \(\varphi\) give zero.
