Voltage, where V = IR, is the product of the current ( I ) of Amperes in an electrical circuit in units of coulombs per second ( q/s ), and resistance ( R ) in Ohms. Likewise, voltage is defined as the amount of energy in Joules ( J ) a coulomb of charge carries within an electric circuit. A coulomb is defined as 6.24*10^18 electrons. In the same way that a baseball thrown by an adult pitcher has more energy than a baseball thrown by a child, a coulomb of electricity will have energy that is determined by the magnitude of the ( + ) and ( – ) charge differential across the circuit.
The simplest of circuit analysis includes a battery ( or some other source of current ) and a single circuit element that resists the flow of electrical charge. Work is done on a resistor when electrons move across it, and the electrons lose a corresponding ability to do work upon passing ( R ). When a coulomb of charge makes a complete trip around an electric circuit, its voltage drop will equal the voltage of the source of current.
In multi-loop circuits, a node is defined as a region where currents “ split up “ and travel from one wire into two or more wires. Current in = Current out, and I1 = I2 + I3 when a current from a single wire travels into two wires. Considerations involving circuit nodes, for the moment, are of no concern.
Let’s assume that source voltage ( Vs ) in the attached image is 12V. Let’s also assume that R1 and R2 have values of 1 Ohm. The total resistance ( Rt ) in a series circuit is the sum of the resistors, and Rt = R1 + R2 = 2R1. Since V = IR, voltage of 12V = ( I )( 2R1 ). The value of the current, therefore, is ( 12V / 2R1 ) = I = 6A. Since 6A of current crosses each 1 Ohm resistor, a voltage drop of Vr = ( 6A )( 1 Ohm ) = 6V occurs at each resistor. This makes intuitive sense as well: The battery has a value of 12V, and two equal 1 Ohm resistors do 6J of work upon each coulomb of charge that passes over them.
Conventional current flows from the ( + ) terminal ( anode ) to the ( – ) terminal ( cathode ) of the circuit. This is symbolic of an electron leaving a high-energy state and moving towards a more stable, lower-energy state. Energy moves from a high to low energy state spontaneously, and a reversal of direction would require an energy input. Therefore, the correct flow of current in the attached diagram would flow in the clockwise direction. What if, however, the wrong assumption were made about the direction of current flow in the circuit?
The sum of all voltage rises and drops in the circuit are +12V – 6V – 6V = 0. This will be the case when the current travels across the circuit in a clockwise direction, but it will also be the case for a hypothetical example in which the electrons travel in the counterclockwise direction. For example, if we take a clockwise journey around the loop, and we begin the journey beneath the battery ( below the negative terminal ), we move from a low to high energy state ( – to + ) as we move upwards across the battery. Thus, a +12V voltage rise is made before two -6V voltage drops are made, and +12V – 6V – 6V = 0. If our journey goes in the counterclockwise direction, and we begin our journey above the + terminal of the battery, -12V + 6V + 6V = 0.
Let’s solve for the value of current with the assumption that current flows in the clockwise direction. If this is the case, the battery MUST be assumed to have a value of +12Vb, because the flow of electrons begins at the ( – ) terminal and rises through the 12V battery towards the ( + ) terminal. Since the total voltage drop is – 12Vr at the resistors, +12Vb – 6Vr – 6Vr = 0, and +12Vb = +12Vr. In terms of I and R, ( + Ib )( Rb ) = ( + Ir )( Rr ). Resistors have positive values only, so solving for the value of current across the resistor gives us [ ( + Ib )( + Rb ) ] / ( + Rr ) = +Ir when done correctly.
Let’s now assume that the current travels in the counterclockwise direction. Since the flow of electricity in the loop begins at the ( + ) terminal of the battery, this assumption MUST include the assumption that a -12V drop occurs as current travels downward across the battery from the + to – terminal. Therefore, -12Vb + 6Vr + 6Vr = 0, and -12Vb = -12Vr. In terms of I and R, ( – Ib )( Rb ) = ( – Ir )( Rr ). Solving for current across the resistors, we have [ ( – Ib )( Rb ) ] / ( Rr ) = ( – Ir ).
In single-loop circuits, the correct current designation is relatively simple to make. Within multi-loop circuits with many circuit elements, however, mistakes can be made. As we will see, Kirchhoff’s Rules will allow us to solve problems even if the incorrect current designation is made. Nonetheless, labeling a drawing or schematic with the correct current designations is crucial to designing electrical circuits that work correctly.
RENAISSANCE PHYSICS 