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Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.
Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop    , in each resistor in [link] .

According to Ohm’s law    , the voltage drop, V size 12{V} {} , across a resistor when a current flows through it is calculated using the equation V = IR size 12{V= ital "IR"} {} , where I size 12{I} {} equals the current in amps (A) and R size 12{R} {} is the resistance in ohms Ω size 12{ left ( %OMEGA right )} {} . Another way to think of this is that V size 12{V} {} is the voltage necessary to make a current I size 12{I} {} flow through a resistance R size 12{R} {} .

So the voltage drop across R 1 size 12{R rSub { size 8{1} } } {} is V 1 = IR 1 size 12{V rSub { size 8{1} } = ital "IR" rSub { size 8{1} } } {} , that across R 2 size 12{R rSub { size 8{2} } } {} is V 2 = IR 2 size 12{V rSub { size 8{2} } = ital "IR" rSub { size 8{2} } } {} , and that across R 3 size 12{R rSub { size 8{3} } } {} is V 3 = IR 3 size 12{V rSub { size 8{3} } = ital "IR" rSub { size 8{3} } } {} . The sum of these voltages equals the voltage output of the source; that is,

V = V 1 + V 2 + V 3 . size 12{V=V rSub { size 8{1} } +V rSub { size 8{2} } +V rSub { size 8{3} } } {}

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation PE = qV size 12{ ital "PE"= ital "qV"} {} , where q size 12{q} {} is the electric charge and V size 12{V} {} is the voltage. Thus the energy supplied by the source is qV size 12{ ital "qV"} {} , while that dissipated by the resistors is

qV 1 + qV 2 + qV 3 . size 12{ ital "qV" rSub { size 8{1} } + ital "qV" rSub { size 8{2} } + ital "qV" rSub { size 8{3} } } {}

Connections: conservation laws

The derivations of the expressions for series and parallel resistance are based on the laws of conservation of energy and conservation of charge, which state that total charge and total energy are constant in any process. These two laws are directly involved in all electrical phenomena and will be invoked repeatedly to explain both specific effects and the general behavior of electricity.

These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus, qV = qV 1 + qV 2 + qV 3 size 12{ ital "qV"= ital "qV" rSub { size 8{1} } + ital "qV" rSub { size 8{2} } + ital "qV" rSub { size 8{3} } } {} . The charge q size 12{q} {} cancels, yielding V = V 1 + V 2 + V 3 size 12{V=V rSub { size 8{1} } +V rSub { size 8{2} } +V rSub { size 8{3} } } {} , as stated. (Note that the same amount of charge passes through the battery and each resistor in a given amount of time, since there is no capacitance to store charge, there is no place for charge to leak, and charge is conserved.)

Now substituting the values for the individual voltages gives

V = IR 1 + IR 2 + IR 3 = I ( R 1 + R 2 + R 3 ) . size 12{V= ital "IR" rSub { size 8{1} } + ital "IR" rSub { size 8{2} } + ital "IR" rSub { size 8{3} } =I \( R rSub { size 8{1} } +R rSub { size 8{2} } +R rSub { size 8{3} } \) } {}

Note that for the equivalent single series resistance R s , we have

V = IR s .

This implies that the total or equivalent series resistance R s of three resistors is R s = R 1 + R 2 + R 3 size 12{R rSub { size 8{s} } =R rSub { size 8{1} } +R rSub { size 8{2} } +R rSub { size 8{3} } } {} .

This logic is valid in general for any number of resistors in series; thus, the total resistance R s of a series connection is

R s = R 1 + R 2 + R 3 + . . . , size 12{R rSub { size 8{s} } =R rSub { size 8{1} } +R rSub { size 8{2} } +R rSub { size 8{3} } + "." "." "." } {}

as proposed. Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up.

Calculating resistance, current, voltage drop, and power dissipation: analysis of a series circuit

Suppose the voltage output of the battery in [link] is 12 . 0 V size 12{"12" "." 0`V} {} , and the resistances are R 1 = 1 . 00 Ω size 12{R rSub { size 8{1} } =1 "." "00" %OMEGA } {} , R 2 = 6 . 00 Ω size 12{R rSub { size 8{2} } =6 "." "00" %OMEGA } {} , and R 3 = 13 . 0 Ω size 12{R rSub { size 8{3} } ="13" "." 0 %OMEGA } {} . (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors.

Strategy and Solution for (a)

The total resistance is simply the sum of the individual resistances, as given by this equation:

Practice Key Terms 9

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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