Physics · Ch 2 — Current Electricity

Resistors in Series and Parallel

2.2.2

Resistors in Series and Parallel

When two or more resistors are connected end to end so that the same current has no alternative path but to flow through every one of them in turn, they are said to be connected in series. Because charge cannot accumulate anywhere in a circuit, the current I passing through resistor $R_1$ must be exactly the same current that passes through $R_2$ and $R_3$ . Since the same current flows through resistors of different resistance, Ohm's law tells us the voltage drop across each one must in general differ: if $V_1, V_2, V_3$ are the voltage drops across $R_1, R_2, R_3$ respectively, then $V_1=IR_1$ , $V_2=IR_2$ , $V_3=IR_3$ , and the supply voltage V must equal the sum of these individual drops:

V = V_1+V_2+V_3 = IR_1+IR_2+IR_3 = I(R_1+R_2+R_3) = IR_S \qquad (2.21, 2.22)

where $R_S$ is the equivalent resistance of the series combination,

R_S = R_1+R_2+R_3 \qquad (2.23)

so several resistors in series simply add up. The equivalent resistance of a series combination is always GREATER than the largest of the individual resistances.

Resistors are instead said to be connected in parallel when they are all connected across the very same two points, so each one experiences the identical potential difference V, but the total current I leaving the battery splits up into separate branch currents $I_1, I_2, I_3$ through each resistor. Conservation of charge again requires

I = I_1+I_2+I_3 \qquad (2.24)

and since the voltage across each resistor is the same, Ohm's law gives $I_1=V/R_1$ , $I_2=V/R_2$ , $I_3=V/R_3$ (2.25). Substituting into (2.24),

I = V\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\right) = \dfrac{V}{R_P}

so the equivalent resistance $R_P$ of a parallel combination satisfies

$\dfrac{1}{R_P} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3} \qquad (2.26)$ …

Figure 2.9Resistors in series

What this figure shows. Panel (a) shows three resistors $R_1$ , $R_2$ , $R_3$ connected end to end in a single loop with a battery, with individual voltage drops $V_1$ , $V_2$ , $V_3$ marked across each and the same current I flowing through all three and through the battery. Panel (b) shows the same circuit replaced by a single equivalent resistor $R_S$ carrying that identical current I, illustrating that a series combination can always be collapsed into on …

Figure 2.10Resistors in parallel

What this figure shows. Panel (a) shows three resistors $R_1$ , $R_2$ , $R_3$ all connected between the same two nodes, across the same battery, with the total current I from the battery splitting into three branch currents $I_1$ , $I_2$ , $I_3$ (one per resistor) that recombine before returning to the battery. Panel (b) shows the same circuit replaced by a single equivalent resistor $R_P$ carrying the same total current I, illustrating that the whole parallel network is equivale …

Misc Example 2.8Series circuit: equivalent resistance and voltage split

Worked out. A $4\ \Omega$ resistor $R_1$ and a $6\ \Omega$ resistor $R_2$ are connected in series to a 24 V battery; the equivalent resistance and the voltage across each resistor are required. Since they are in series, $R_{eq} = 4+6 = 10\ \Omega$ , giving current $I = V/R_{eq} = 24/10 = 2.4$ A. The voltage across the $4\ \Omega$ resistor is $V_1 = IR_1 = 2.4\times4 = 9.6$ V, and across the $6\ \Omega$ resistor is $V_2 = IR_2 = 2.4\times6 = 14.4$ V; as a check, $9.6+14.4 = 24$ V, exactly the …

Misc Example 2.9Parallel circuit: equivalent resistance and branch currents

Worked out. A $4\ \Omega$ resistor $R_1$ and a $6\ \Omega$ resistor $R_2$ are connected in parallel to a 24 V battery; the equivalent resistance and the currents $I$ , $I_1$ , $I_2$ are required. Since they are in parallel, $1/R_P = 1/4+1/6 = 5/12$ , so $R_P = 12/5 = 2.4\ \Omega$ . Because both resistors share the same 24 V, $I_1 = 24/4 = 6$ A and $I_2 = 24/6 = 4$ A, and the total current from the battery is their sum, $I = I_1+I_2 = 6+4 = 10$ …

Misc Example 2.10Finding two resistances from their series and parallel equivalents

Worked out. Two resistors, when connected in series, give an equivalent resistance of $15\ \Omega$ , and when connected in parallel give $56/15\ \Omega$ ; their individual values are required. From the series condition, $R_1+R_2 = 15$ . From the parallel condition, $R_1R_2/(R_1+R_2) = 56/15$ , so $R_1R_2 = (56/15)\times15 = 56$ . Substituting $R_2 = 15-R_1$ into the product gives the quadratic $R_1^2-15R_1+56=0$ , which factorises to give $R_1 = 8\ \Omega$ or $R_1 = 7\ \Omega$ . Correspondingly, when $R_1=8\ \Omega$ , $R_2=7\ \Omega$ , and vice versa -- the two resistors are $7\ \Omega$ and $8\ \Omega$ , and the problem cannot distinguish which is 'first' since both series and parallel combina …

Misc Example 2.11Reducing a three-section ladder network by parallel groups

Worked out. A network runs from A to B through three sections, each made of two parallel resistors: the first section has two $2\ \Omega$ resistors in parallel, the second has two $4\ \Omega$ resistors in parallel, and the third has two $6\ \Omega$ resistors in parallel, with the three sections then connected in series between A and B. Each parallel pair of equal resistors R reduces to R/2, so the three sections reduce to $2/2=1\ \Omega$ , $4/2=2\ \Omega$ and $6/2=3\ \Omega$ respectively. These three reduced values are now in series along the A-to-B path, so the total equivalent resistance is $R = 1+2+3 = 6\ \Omega$ . This problem is a good illustration of the general strategy for any resistor network: repeatedly collapse obvious series or parallel groups from …

Misc Example 2.12Using symmetry to eliminate a bridging resistor

Worked out. Five resistors connect four points a, b, c, d: a $5\ \Omega$ resistor bridges c and d, while four $1\ \Omega$ resistors form the outer loop a-c, c-b, a-d and d-b; the equivalent resistance between a and b is required. Because all four outer-loop resistors are equal ( $1\ \Omega$ each), a current entering at a splits exactly equally into the a-c and a-d branches, so points c and d must be at exactly the same potential. With no potential difference between c and d, no current at all flows through the $5\ \Omega$ bridging resistor, so it can simply be removed from the circuit without changing anything. What remains is two parallel paths from a to b, each made of two $1\ \Omega$ resistors in series (giving $2\ \Omega$ per path), and those two $2\ \Omega$ paths in parallel give an equivalent resistance of $R_{eq} = 1\ \Omega$ between a and b. This is the classic bridge-symmetry trick: whenever a bridge element sits between two points …

Misc ~note-appliances-parallelWhy household appliances are wired in parallel

Worked out. A short practical note: household appliances are always connected in parallel across the mains supply (rather than in series) precisely because a parallel connection means every appliance gets the same full supply voltage and operates independently -- so switching one appliance off does not interrupt the current to, or change the voltage across, any of the others. …

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