Physics · Ch 11 — Electric Current Through Conductors

Parallel Combination of Resistors

11.8.2.2

Parallel Combination of Resistors

In a PARALLEL combination, the resistors are instead connected between the SAME two electrical (junction) points, so that each resistor forms its own individual path between those points and the SAME voltage $V$ is applied across every resistor in the group (Fig. 11.10). The total current $I$ supplied splits between the branches -- $I_1$ through $R_1$ , $I_2$ through $R_2$ , and so on -- while the voltage stays common:

I=I_1+I_2\qquad\text{--- (11.27)}

Applying Ohm's law separately to each branch (since each sees the full voltage $V$ ):

V=I_1R_1\ \Rightarrow\ I_1=\frac{V}{R_1}\qquad\text{--- (11.28a)}\qquad\qquad V=I_2R_2\ \Rightarrow\ I_2=\frac{V}{R_2}\qquad\text{--- (11.28b)}

Substituting into Eq. (11.27),

I=\frac{V}{R_1}+\frac{V}{R_2}

If this same total current is written as $I=V/R_p$ for a single equivalent resistor $R_p$ replacing the parallel pair,

\frac{V}{R_p}=\frac{V}{R_1}+\frac{V}{R_2}\ \Rightarrow\ \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}\qquad\text{--- (11.29)}

For $n$ resistors $R_1,R_2,\dots,R_n$ all connected in parallel, this generalises to the reciprocal-sum rule

$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\dots+\frac{1}{R_n}=\sum_{i=1}^{n}\frac{1}{R_i}\qquad\text{--- (11.30)}$ …

Figure 11.10Two resistors in parallel combination

What this figure shows. A circuit diagram showing two resistors R1 and R2 connected between the SAME pair of electrical junction points (nodes), so each resistor forms its own separate individual path between those two common points, and the same voltage V appears across both of them. The total current I entering the parallel combination is shown splitting into two branch currents I1 (through R1) and I2 (through R2), which recombine to the total I again after passing through their respective resistors, illustrating I = …

Misc Ex.5Example 11.5 -- Total resistance and total current for a mixed series-parallel resistor circuit

Worked out. A circuit with R1=3Ω, R2=6Ω and R3=5Ω and a 14 V supply, where R1 and R2 are combined in parallel and that combination is then placed in series with R3. The worked solution first finds the parallel combination Rp = (R1×R2)/(R1+R2) = (3×6)/9 = 2Ω, then adds it in series with R3 to get the total resistance RT = Rp+R3 = 2+5 = 7Ω, and finally applies Ohm's law I = V/RT = 14/7 = 2A to find the total current drawn from the 14 V supply. …

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