Difference between revisions of "Calculate Series and Parallel Resistance"
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===Series Resistance=== | ===Series Resistance=== | ||
#What it is. Series resistance is simply connecting the "out" side of one resistor to the "in" side of another in a circuit. Each additional resistor placed in a circuit adds to the total resistance of that circuit. | #What it is. Series resistance is simply connecting the "out" side of one resistor to the "in" side of another in a circuit. Each additional resistor placed in a circuit adds to the total resistance of that circuit. | ||
| − | #*The formula for calculating a total of ''n'' number of resistors wired in series is: | + | #*The formula for calculating a total of ''n'' number of resistors wired in series is: R<sub>eq</sub> = R<sub>1</sub> + R<sub>2</sub> + .... R<sub>n</sub> <br> That is, all the series resistor values are simply added. For example, consider finding the equivalent resistance in the image below |
| − | #*In this example,<br> R<sub>1</sub> = 100 Ω and R<sub>2</sub> = 300Ω are wired in series. R<sub>eq</sub> = 100 Ω + 300 Ω = 400 Ω | + | #*In this example,<br> R<sub>1</sub> = 100 Ω and R<sub>2</sub> = 300Ω are wired in series. R<sub>eq</sub> = 100 Ω + 300 Ω = 400 Ω |
===Parallel Resistance=== | ===Parallel Resistance=== | ||
#What it is. Parallel resistance is when the "in" side of 2 or more resistors are connected, and the "out" side of those resistors are connected. | #What it is. Parallel resistance is when the "in" side of 2 or more resistors are connected, and the "out" side of those resistors are connected. | ||
| − | #*The equation for combining ''n'' resistors in parallel is: | + | #*The equation for combining ''n'' resistors in parallel is:R<sub>eq</sub> = 1/{(1/R<sub>1</sub>)+(1/R<sub>2</sub>)+(1/R<sub>3</sub>)..+(1/R<sub>n</sub>)} |
| − | #*Here is an example, given R<sub>1</sub> = 20 Ω, R<sub>2</sub> = 30 Ω, and R<sub>3</sub> = 30 Ω. | + | #*Here is an example, given R<sub>1</sub> = 20 Ω, R<sub>2</sub> = 30 Ω, and R<sub>3</sub> = 30 Ω. |
| − | #*The total equivalent resistance for all 3 resistors in parallel is: | + | #*The total equivalent resistance for all 3 resistors in parallel is:R<sub>eq</sub> = 1/{(1/20)+(1/30)+(1/30)} = 1/{(3/60)+(2/60)+(2/60)} = 1/(7/60)=60/7 Ω = approximately 8.57 Ω. |
===Combined Series and Parallel Circuits=== | ===Combined Series and Parallel Circuits=== | ||
#What it is. A combined network is any combination of series and parallel circuits wired together. Consider finding the equivalent resistance of the network shown below. | #What it is. A combined network is any combination of series and parallel circuits wired together. Consider finding the equivalent resistance of the network shown below. | ||
| − | #* We see the resistors R<sub>1</sub> and R<sub>2</sub> are connected in series. So their equivalent resistance (let us denote it by R<sub>s</sub>) is: | + | #* We see the resistors R<sub>1</sub> and R<sub>2</sub> are connected in series. So their equivalent resistance (let us denote it by R<sub>s</sub>) is: R<sub>s</sub> = R<sub>1</sub> + R<sub>2</sub> = 100 Ω + 300 Ω = 400 Ω. |
| − | #* Next, we see the resistors R<sub>3</sub> and R<sub>4</sub> are connected in parallel. So their equivalent resistance (let us denote it by R<sub>p1</sub>) is: | + | #* Next, we see the resistors R<sub>3</sub> and R<sub>4</sub> are connected in parallel. So their equivalent resistance (let us denote it by R<sub>p1</sub>) is: R<sub>p1</sub> = 1/{(1/20)+(1/20)} = 1/(2/20)= 20/2 = 10 Ω |
| − | #* Then we see the resistors R<sub>5</sub> and R<sub>6</sub> are also connected in parallel. So their equivalent resistance (let us denote it by R<sub>p2</sub>) is: | + | #* Then we see the resistors R<sub>5</sub> and R<sub>6</sub> are also connected in parallel. So their equivalent resistance (let us denote it by R<sub>p2</sub>) is: R<sub>p2</sub> = 1/{(1/40)+(1/10)} = 1/(5/40) = 40/5 = 8 Ω |
| − | #* So now we have a circuit with the resistors R<sub>s</sub>, R<sub>p1</sub>, R<sub>p2</sub> and R<sub>7</sub> connected in series. These can now simply be added to get the equivalent resistance R<sub>7</sub> of the network given to us originally. | + | #* So now we have a circuit with the resistors R<sub>s</sub>, R<sub>p1</sub>, R<sub>p2</sub> and R<sub>7</sub> connected in series. These can now simply be added to get the equivalent resistance R<sub>7</sub> of the network given to us originally.R<sub>eq</sub> = 400 Ω + 20Ω + 8 Ω = 428 Ω. |
==Some Facts== | ==Some Facts== | ||
