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5.1 Resistors in series and parallel  (Page 5/17)

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Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in [link] . Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The process is more time consuming than difficult.

The diagram has a set of five circuits. The first circuit has a combination of seven resistors in series and parallel combinations. It has a resistor R sub one in series with a set of three resistors R sub two, R sub three, and R sub four in parallel and connected in series with a combination of resistors R sub five and R sub six, which are parallel. A resistor R sub seven is connected in parallel to R sub one and the voltage source. The second circuit calculates combinations of all parallel resistors in circuit one and replaces them with their equivalent resistance. It has a resistor R sub one in series with R sub p and R sub p prime. A resistor R sub seven is connected in parallel to R sub one and the voltage source. The third circuit takes the combination of the series resistors R sub p and R sub p prime and replaces it with R sub s. It has a resistor R sub one in series with R sub s. A resistor R sub seven is connected in parallel to R sub s and the voltage source. The fourth circuit shows a parallel combination of R sub seven and R sub s are calculated and replaced by R sub p double prime. The circuit now has a series combination voltage source, R sub one and R sub p double prime. The fifth circuit shows the final equivalent of the first circuit. It has a voltage source connecting across a final equivalent resistance R sub s prime.
This combination of seven resistors has both series and parallel parts. Each is identified and reduced to an equivalent resistance, and these are further reduced until a single equivalent resistance is reached.

The simplest combination of series and parallel resistance, shown in [link] , is also the most instructive, since it is found in many applications. For example, R 1 size 12{R rSub { size 8{1} } } {} could be the resistance of wires from a car battery to its electrical devices, which are in parallel. R 2 size 12{R rSub { size 8{1} } } {} and R 3 size 12{R rSub { size 8{1} } } {} could be the starter motor and a passenger compartment light. We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects, as the next example indicates.

Calculating resistance, IR size 12{ ital "IR"} {} Drop, current, and power dissipation: combining series and parallel circuits

[link] shows the resistors from the previous two examples wired in a different way—a combination of series and parallel. We can consider R 1 size 12{R rSub { size 8{1} } } {} to be the resistance of wires leading to R 2 size 12{R rSub { size 8{2} } } {} and R 3 size 12{R rSub { size 8{3} } } {} . (a) Find the total resistance. (b) What is the IR size 12{ ital "IR"} {} drop in R 1 size 12{R rSub { size 8{1} } } {} ? (c) Find the current I 2 size 12{I rSub { size 8{2} } } {} through R 2 size 12{R rSub { size 8{2} } } {} . (d) What power is dissipated by R 2 size 12{R rSub { size 8{2} } } {} ?

Circuit diagram in which a battery of twelve point zero volts is connected to a combination of three resistors. Resistors R sub two and R sub three are connected in parallel to each other, and their combination is connected in series to resistor R sub one. R sub one has a resistance of one point zero zero ohms, R sub two has a resistance of six point zero zero ohms, and R sub three has a resistance of thirteen point zero ohms.
These three resistors are connected to a voltage source so that R 2 size 12{R rSub { size 8{2} } } {} and R 3 size 12{R rSub { size 8{3} } } {} are in parallel with one another and that combination is in series with R 1 size 12{R rSub { size 8{1} } } {} .

Strategy and Solution for (a)

To find the total resistance, we note that R 2 size 12{R rSub { size 8{2} } } {} and R 3 size 12{R rSub { size 8{3} } } {} are in parallel and their combination R p size 12{R rSub { size 8{p} } } {} is in series with R 1 size 12{R rSub { size 8{1} } } {} . Thus the total (equivalent) resistance of this combination is

R tot = R 1 + R p . size 12{R rSub { size 8{"tot"} } =R rSub { size 8{1} } +R rSub { size 8{p} } } {}

First, we find R p size 12{R rSub { size 8{p} } } {} using the equation for resistors in parallel and entering known values:

1 R p = 1 R 2 + 1 R 3 = 1 6 . 00 Ω + 1 13 . 0 Ω = 0 . 2436 Ω . size 12{ { {1} over {R rSub { size 8{p} } } } = { {1} over {R rSub { size 8{2} } } } + { {1} over {R rSub { size 8{3} } } } = { {1} over {6 "." "00" %OMEGA } } + { {1} over {"13" "." 0 %OMEGA } } = { {0 "." "2436"} over { %OMEGA } } } {}

Inverting gives

R p = 1 0 . 2436 Ω = 4 . 11 Ω . size 12{R rSub { size 8{p} } = { {1} over {0 "." "2436"} } %OMEGA =4 "." "11" %OMEGA } {}

So the total resistance is

R tot = R 1 + R p = 1 . 00 Ω + 4 . 11 Ω = 5 . 11 Ω . size 12{R rSub { size 8{"tot"} } =R rSub { size 8{1} } +R rSub { size 8{p} } =1 "." "00" %OMEGA +4 "." "11 " %OMEGA =5 "." "11 " %OMEGA } {}

Discussion for (a)

The total resistance of this combination is intermediate between the pure series and pure parallel values ( 20.0 Ω and 0.804 Ω , respectively) found for the same resistors in the two previous examples.

Strategy and Solution for (b)

To find the IR size 12{ ital "IR"} {} drop in R 1 size 12{R rSub { size 8{1} } } {} , we note that the full current I size 12{I} {} flows through R 1 size 12{R rSub { size 8{1} } } {} . Thus its IR size 12{ ital "IR"} {} drop is

V 1 = IR 1 . size 12{V rSub { size 8{1} } = ital "IR" rSub { size 8{1} } } {}

We must find I size 12{I} {} before we can calculate V 1 size 12{V rSub { size 8{1} } } {} . The total current I size 12{I} {} is found using Ohm’s law for the circuit. That is,

I = V R tot = 12 . 0 V 5 . 11 Ω = 2 . 35 A . size 12{I= { {V} over {R rSub { size 8{"tot"} } } } = { {"12" "." 0" V"} over {5 "." "11 " %OMEGA } } =2 "." "35"" A"} {}

Entering this into the expression above, we get

V 1 = IR 1 = ( 2 . 35 A ) ( 1 . 00 Ω ) = 2 . 35 V . size 12{V rSub { size 8{1} } = ital "IR" rSub { size 8{1} } = \( 2 "." "35"" A" \) \( 1 "." "00" %OMEGA \) =2 "." "35"" V"} {}

Discussion for (b)

The voltage applied to R 2 size 12{R rSub { size 8{2} } } {} and R 3 size 12{R rSub { size 8{3} } } {} is less than the total voltage by an amount V 1 size 12{V rSub { size 8{1} } } {} . When wire resistance is large, it can significantly affect the operation of the devices represented by R 2 size 12{R rSub { size 8{2} } } {} and R 3 size 12{R rSub { size 8{3} } } {} .

Strategy and Solution for (c)

To find the current through R 2 size 12{R rSub { size 8{2} } } {} , we must first find the voltage applied to it. We call this voltage V p size 12{V rSub { size 8{p} } } {} , because it is applied to a parallel combination of resistors. The voltage applied to both R 2 size 12{R rSub { size 8{2} } } {} and R 3 size 12{R rSub { size 8{3} } } {} is reduced by the amount V 1 size 12{V rSub { size 8{1} } } {} , and so it is
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Source:  OpenStax, General physics ii phy2202ca. OpenStax CNX. Jul 05, 2013 Download for free at http://legacy.cnx.org/content/col11538/1.2
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