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This lesson will teach you how to calculate all the electrical unknowns of a DC Parallel Circuit, and is a companion to the lesson “Calculating the Electrical Unknowns of a DC Series Circuit.” We will examine the three laws of a parallel circuit, which will help explain how this type of circuit works, and Kirchhoff’s Current Law (KCL), which will help explain current flows in and out of a junction.

For Reference, Review, and PracticeFigure 1. Four Automotive Bulb Filaments in a Parallel Circuit

In order to understand completely the function of a DC Parallel Circuit, all the electrical unknowns must be calculated mathematically. A Parallel Circuit is a complete circuit that has more than one path for current flow. The separate paths, which split and meet at junction points, are called branches or legs. See Figure 1 for a depiction of four automotive bulb filaments in a Parallel Circuit. Before we begin our analysis of this circuit, we must understand the symbols of its factors. Based upon Ohm’s Law,

Vis for voltage (measured in volts),Ris for resistance (measured in ohms, and the symbol is Ω),Iis for current flow (measured in amperes), andPis for power (measured in watts). See the companion lesson “Applying Ohm’s Law” for further instruction.In this automotive application with four bulb filaments in a Parallel Circuit, the source voltage or the Total Voltage (V

_{T}) is 12V DC. In order to understand this Parallel Circuit with four resistive elements, there are 15 unknowns, which must be calculated mathematically. We will use subscripts to help keep track of all the circuit elements. The 15 circuit elements or unknowns are as follows:_{T}= Total Resistance_{T}= Total Current flow_{T}= Total Power dissipated_{R}_{1}= Voltage Drop across R_{1}_{R}_{2}= Voltage Drop across R_{2}_{R}_{3}= Voltage Drop across R_{3}_{R}_{4}= Voltage Drop across R_{4}_{R}_{1}= Current flow through R_{1}_{R}_{2}= Current flow through R_{2}_{R}_{3}= Current flow through R_{3}_{R}_{4}= Current flow through R_{4}_{R}_{1}= Power dissipated by R_{1}_{R}_{2}= Power dissipated by R_{2}_{R}_{3}= Power dissipated by R_{3}_{R}_{4}= Power dissipated by R_{4}When there are only two resistors in a Parallel Circuit, we can calculate Total Resistance (R

_{T}) like this:R

_{1}x R_{2}R

_{T}= R_{1}+ R_{2}## The Three Laws of a Parallel Circuit

The three laws that govern a Parallel Circuit are as follows:

Law #1– The total resistance of a Parallel Circuit is always less than that of the smallest resistance in the circuit.Law #2– In a Parallel Circuit, the voltage is the same across each leg.Law #3– The sum of the individual currents in each leg of a Parallel Circuit will equal the total current flow of that circuit.## Kirchhoff’s Voltage and Current Laws

Kirchhoff’s Voltage Law (KVL): Applies to Series CircuitsThe voltage around any closed circuit is equal to the sum (total voltage) of the voltage drops across the resistances in the closed loop.

The mathematics of this law is expressed in the equations below:

∑ ∆V = 0 (Note: The Greek letter Sigma “∑” means “summation in mathematics” and the

Greek letter Delta ∆ means “a change in”).

V

_{1+}V_{2+}V_{3}= 0V

_{T}= V_{1+}V_{2+}V_{3+}. . . + V_{n}## Kirchhoff’s Current Law (KCL): Applies to Parallel Circuits

The algebraic sum of all currents entering and exiting a node must equal zero.

The mathematics of this law is expressed in the equation below:

∑ I

_{In}= ∑ I_{Out}= 0I

_{1+}I_{2+}I_{3 –}I_{4}– I_{5}– I_{6}= 0## Formulas to Know

(Note: These equations are listed in the order they should be used.)

Total Resistance: R

_{T}= 1(1/R_{1}+1/R_{2+}1/R_{3}+1/R_{4})Current flow through R

_{1}: I_{R}_{1}= V_{R}_{1}/R_{1}Total Current: I

_{T}= V_{T}/R_{T}or I_{T}= I_{R}_{1}+ I_{R}_{2}+I_{R}_{3}+I_{R}_{4}Current flow through R

_{2}: I_{R}_{2}= V_{R}_{2}/R_{2}Current flow through R

_{3}: I_{R}_{3}= V_{R}_{3}/R_{3}Total Power: P

_{T}= P_{R}_{1}+P_{R}_{2}+P_{R}_{3}+P_{R}_{4}. . .Current flow through R

_{4}: I_{R}_{4}= V_{R}_{4}/R_{4}Voltage Drop across R

_{1}: V_{R}_{1}= I_{R}_{1}x R_{1}Current flow through R

_{4}: I_{R}_{4}= V_{R}_{4}/R_{4}Voltage Drop across R

_{2}: V_{R}_{2}= I_{R}_{2}x R_{2}Power dissipated across R

_{1}: P_{R}_{1}= I_{R}_{1}_{x }R_{1}Voltage Drop across R

_{3}: V_{R}_{3}= I_{R}_{3}x R_{3}Power dissipated across R

_{2}: P_{R}_{2}= I_{R}_{2}_{x }R_{2}Voltage Drop across R

_{4:}V_{R}_{4}= I_{R}_{4}x R_{4}Power dissipated across R

_{3}: P_{R}_{3}= I_{R}_{3}_{x }R_{3}Power dissipated across R

_{4}: P_{R}_{4}= I_{R}_{4}_{x}R_{4}R

_{T}= V_{T}/I_{T}R

_{T}=(R_{1}x R_{2})/(R_{1}+ R_{2})learn by doing

## Parallel Circuits