{"id":109,"date":"2021-06-20T23:15:30","date_gmt":"2021-06-20T23:15:30","guid":{"rendered":"https:\/\/mc.scsiraidguru.com\/?page_id=109"},"modified":"2024-01-09T21:15:25","modified_gmt":"2024-01-09T21:15:25","slug":"6-resistors-serial-and-parallel","status":"publish","type":"page","link":"https:\/\/mc.scsiraidguru.com\/index.php\/electrical-engineering\/resistor\/6-resistors-serial-and-parallel\/","title":{"rendered":"6 resistors: serial and parallel"},"content":{"rendered":"\t\t
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The j2 to j3 leg we add a zero resistor<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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I will break down the circuit<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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We will join the j1 to j2 series resistors<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Now combine J1 to J3 resistors in series : 670 + 0 = 670\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Now combine j1 and j3 in parallel.
1\/R = 1\/670 + 1\/1000
R = 401.2<\/p><\/div><\/div><\/div><\/div><\/div><\/div><\/div><\/div><\/section>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Add what is left in series<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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now we can get the current of the circuit : I o <\/sub> = 9V \/ 2400\u03a9<\/p>

I o <\/sub>= 3.75mA<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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We can now figure out the voltage drops<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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V 1000 <\/sub> = 1k\u03a9 * 3.75mA = 3.75V
V 401 <\/sub> = 401.2\u03a9 * 3.75mA = 1.5V
V 1000 <\/sub> = 1k\u03a9 * 3.75mA = 3.75V<\/p>

Series resistors have same current but different voltages<\/p><\/div><\/div><\/div><\/div><\/div><\/div><\/div><\/div><\/section>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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j1 to J3 is made up of two resistors 670\u03a9 and 1k\u03a9.
V 401 <\/sub> = 401.2\u03a9 * 3.75mA = 1.5V<\/p>

So this voltage goes through both resistors
V 670 <\/sub> = 1.5V and V1000 <\/sub> = 1.5V<\/p>

\u00a0I670 <\/sub> = 1.5V \/ 670\u03a9 = 2.24mA
I1000 <\/sub> = 1.5V \/ 1000 = 1.5mA<\/p><\/div><\/div><\/div><\/div><\/div><\/div><\/div><\/div><\/section>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Zero resistor has no voltage or current.\u00a0 It was a place holder.<\/p>

Resistors in parallel have same voltages but different currents<\/p><\/div><\/div><\/div><\/div><\/div><\/div><\/div><\/div><\/section>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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The current from j1 to j2<\/p>

\u00a0I2000 <\/sub> = 1.5V\/2000\u03a9 = 0.75mA
I1000<\/sub> =1.5V \/ 1000\u03a9 = 1.5mA<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Last redraw<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Final circuit<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"\"\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"

The j2 to j3 leg we add a zero resistor I will break down the circuit We will join the j1 to j2 series resistors Now combine J1 to J3 resistors in series : 670 + 0 = 670\u00a0 Now combine j1 and j3 in parallel.1\/R = 1\/670 + 1\/1000R = 401.2 Add what is 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