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12.1 Resistors in series and parallel  (Page 4/17)

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Strategy and Solution for (a)

The total resistance for a parallel combination of resistors is found using the equation below. Entering known values gives

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

Thus,

1 R p = 1.00 Ω + 0 . 1667 Ω + 0 . 07692 Ω = 1 . 2436 Ω . size 12{ { {1} over {R rSub { size 8{p} } } } = { {1 "." "00"} over { %OMEGA } } + { {0 "." "167"} over { %OMEGA } } + { {0 "." "0769"} over { %OMEGA } } = { {1 "." "244"} over { %OMEGA } } } {}

(Note that in these calculations, each intermediate answer is shown with an extra digit.)

We must invert this to find the total resistance R p size 12{R rSub { size 8{p} } } {} . This yields

R p = 1 1 . 2436 Ω = 0 . 8041 Ω . size 12{R rSub { size 8{p} } = { {1} over {1 "." "2436"} } %OMEGA =0 "." "8041 " %OMEGA } {}

The total resistance with the correct number of significant digits is R p = 0 . 804 Ω . size 12{R rSub { size 8{p} } =0 "." "804" %OMEGA } {}

Discussion for (a)

R p is, as predicted, less than the smallest individual resistance.

Strategy and Solution for (b)

The total current can be found from Ohm’s law, substituting R p size 12{R rSub { size 8{p} } } {} for the total resistance. This gives

I = V R p = 12.0 V 0.8041 Ω = 14 . 92 A . size 12{I= { {V} over {R rSub { size 8{p} } } } = { {"12" "." 0" V"} over {0 "." "804 " %OMEGA } } ="14" "." "92"" A"} {}

Discussion for (b)

Current I size 12{I} {} for each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallel connections has a smaller total resistance than the resistors connected in series.

Strategy and Solution for (c)

The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus,

I 1 = V R 1 = 12 . 0 V 1 . 00 Ω = 12 . 0 A . size 12{I rSub { size 8{1} } = { {V} over {R rSub { size 8{1} } } } = { {"12" "." 0" V"} over {1 "." "00 " %OMEGA } } ="12" "." 0" A"} {}

Similarly,

I 2 = V R 2 = 12 . 0 V 6 . 00 Ω = 2 . 00 A size 12{I rSub { size 8{2} } = { {V} over {R rSub { size 8{2} } } } = { {"12" "." 0" V"} over {6 "." "00 " %OMEGA } } =2 "." "00"" A"} {}

and

I 3 = V R 3 = 12 . 0 V 13 . 0 Ω = 0 . 92 A . size 12{I rSub { size 8{3} } = { {V} over {R rSub { size 8{3} } } } = { {"12" "." 0" V"} over {"13" "." "0 " %OMEGA } } =0 "." "92"" A"} {}

Discussion for (c)

The total current is the sum of the individual currents:

I 1 + I 2 + I 3 = 14 . 92 A . size 12{I rSub { size 8{1} } +I rSub { size 8{2} } +I rSub { size 8{3} } ="14" "." "92"" A"} {}

This is consistent with conservation of charge.

Strategy and Solution for (d)

The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use P = V 2 R size 12{P= { {V rSup { size 8{2} } } over {R} } } {} , since each resistor gets full voltage. Thus,

P 1 = V 2 R 1 = ( 12 . 0 V ) 2 1 . 00 Ω = 144 W . size 12{P rSub { size 8{1} } = { {V rSup { size 8{2} } } over {R rSub { size 8{1} } } } = { { \( "12" "." 0" V" \) rSup { size 8{2} } } over {1 "." "00 " %OMEGA } } ="144"" W"} {}

Similarly,

P 2 = V 2 R 2 = ( 12 . 0 V ) 2 6 . 00 Ω = 24 . 0 W size 12{P rSub { size 8{2} } = { {V rSup { size 8{2} } } over {R rSub { size 8{2} } } } = { { \( "12" "." 0" V" \) rSup { size 8{2} } } over {6 "." "00 " %OMEGA } } ="24" "." 0" W"} {}

and

P 3 = V 2 R 3 = ( 12 . 0 V ) 2 13 . 0 Ω = 11 . 1 W . size 12{P rSub { size 8{3} } = { {V rSup { size 8{2} } } over {R rSub { size 8{3} } } } = { { \( "12" "." 0" V" \) rSup { size 8{2} } } over {"13" "." "0 " %OMEGA } } ="11" "." 1" W"} {}

Discussion for (d)

The power dissipated by each resistor is considerably higher in parallel than when connected in series to the same voltage source.

Strategy and Solution for (e)

The total power can also be calculated in several ways. Choosing P = IV size 12{P= ital "IV"} {} , and entering the total current, yields

P = IV = ( 14.92 A ) ( 12.0 V ) = 179 W . size 12{P= ital "IV"= \( "14" "." "92"" A" \) \( "12" "." 0" V" \) ="179" "." 1" W"} {}

Discussion for (e)

Total power dissipated by the resistors is also 179 W:

P 1 + P 2 + P 3 = 144 W + 24 . 0 W + 11 . 1 W = 179 W . size 12{P rSub { size 8{1} } +P rSub { size 8{2} } +P rSub { size 8{3} } ="144"" W"+"24" "." 0" W"+"11" "." 1" W"="179"" W"} {}

This is consistent with the law of conservation of energy.

Overall Discussion

Note that both the currents and powers in parallel connections are greater than for the same devices in series.

Major features of resistors in parallel

  1. Parallel resistance is found from 1 R p = 1 R 1 + 1 R 2 + 1 R 3 + . . . size 12{ { {1} over {R rSub { size 8{p} } } } = { {1} over {R rSub { size 8{1} } } } + { {1} over {R rSub { size 8{2} } } } + { {1} over {R rSub { size 8{3} } } } + "." "." "." } {} , and it is smaller than any individual resistance in the combination.
  2. Each resistor in parallel has the same full voltage of the source applied to it. (Power distribution systems most often use parallel connections to supply the myriad devices served with the same voltage and to allow them to operate independently.)
  3. Parallel resistors do not each get the total current; they divide it.

Combinations of series and parallel

More complex connections of resistors are sometimes just combinations of series and parallel. These are commonly encountered, especially when wire resistance is considered. In that case, wire resistance is in series with other resistances that are in parallel.

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Source:  OpenStax, Introductory physics - for kpu phys 1100 (2015 edition). OpenStax CNX. May 30, 2015 Download for free at http://legacy.cnx.org/content/col11588/1.13
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