Applied Science Department Cheryl Dellai Physics Instructor

A Method for Resolving Series/Parallel Circuits

by Cheryl Kay Dellai

Applied Science Department, Glendale Community College, Glendale, AZ

In any study of electrical circuits beginning students have problems correctly identifying and calculating the relevant values. The more complex the circuit becomes the harder it is. The method I use addresses these problems.

This is a four step method based on reducing everything to a labeled table. When a student first starts resolving circuits, I also require redrawing of the circuit at each step in the process.

Find capacitance C, charge Q, electrical potential difference (voltage) V, and electrical potential energy.

Step One

Label all junctions in the circuit (Figure 1). In the case above there are 6 junctions (a,b,c,d,e,f)

Step Two

Identify the deepest levels of the circuit by tracing the branches. In the case above Branch cd is the deepest branch and must be resolved first. Branch be is second; Branch af is third.

Step Three

Within the deepest branch circle the series and parallel pieces and label them. Write the branch components and arrangement (series/parallel) into a table (Table I) as shown. In the Figure 1 Branch cd is a parallel arrangement. Each arm has more than one capacitor. These arms must be treated first.

Right Arm (See figure 2)

Capacitors 3.00 mF & 4.00 mF are in series and must reduced to Capacitor A (1.71 mF). )

Left Arm (See figure 2)

Capacitors 9.00 mF & 8.00 mF are in series and must reduced to Capacitor B (4.24 mF).

Redraw the circuit after each reduction.

Capacitors A & B are in parallel in Figure 2 Branch cd. We reduce them to Capacitor C (5.95 mF). At this point we have eliminated Branch cd and can move on to Branch be and Branch af.

In Figure 3 Branch be is a parallel arrangement. The right arm has more than one capacitor. This Branch be arm must be treated first. At that point Branch be is part of a parallel arrangement in Branch af.

Branch be

Right Arm

Capacitors C 5.95 mF & 2.00 mF are in series and must reduced to Capacitor D (1.50 mF).

Center Arm Capacitor 7.00 mF

Branch af

Left Arm Capacitor 6.00 mF

Capacitors D, 6.00 mF & 7.00 mF are in parallel in Branch af. We reduce them to Capacitor E (14.5 mF). At this point we have eliminated Branch af.

In the case below (Fig 5) we have a series arrangement. Capacitors E 14.5, 5.00 mF & 1.00 mF are in series and must reduced to Capacitor F (0.79 mF) in Figure 6.

Step Four

The circuit now has an equivalent capacitance. We need to solve for total charge, and total potential energy using the given electrical potential difference.

Now we work our way back up the table.

If a branch is in parallel, the electric potential difference (voltage)is the same, and charge is additive ( q_total = S q_i).

If a branch is in series, the charge is the same, and electrical potential difference (voltage) is additive ( V_total = S V_i ).

Potential energy is always additive ( U_total = S U_i).

We will breakdown each equivalent capacitor as we move back up the table.

Capacitor F is a series combination. Thus, the charge on capacitors 1.00 mF, 5.00 mF and Capacitor E (14.50 mF) is the total charge (78.80 mC), and the electric potential differences (voltages) must add up to the total electrical potential difference of 100 volts (voltage). If it does not, a mistake was made in calculating the capacitance. That is an internal check this table provides.

We must then breakdown Capacitor E. Capacitor E is a parallel combination. Thus, the electrical potential difference (voltage) of 5.44 volts on E is the same electrical potential difference as (voltage) on capacitors 6.00 mF, 7.00 mF, and Capacitor D (1.50 mF). The charge on each of these capacitors must add up to the charge on Capacitor E. If it does not, a mistake was made in calculating the capacitance.

We repeat this process until we finish the table.

Students can complete a table like this in 10 minutes.

Table I

	Capacitance	Charge	Voltage	Energy
	mF	mC	Volts	mJ
Series	3.00		0.78	0.92
A	4.00	2.34	0.59	0.69
Series	9.00		0.64	1.86
B	8.00	5.79	0.72	2.10
Parallel	A 1.71	2.34		1.60
C	B 4.24	5.79	1.37	3.96
Series	C 5.95		1.37	5.56
D	2.00	8.14	4.07	16.55
Parallel	D 1.50	8.14		22.11
E	7.00	38.05	5.44	103.42
	6.00	32.62		88.65
Series	E 14.50		5.44	214.18
F	1.00	78.80	78.80	3104.99
	5.00		15.76	621.00
Total	0.79	78.80	100.00	3940.17

This is an equivalent resistance circuit and table. Find Resistance R, current i, electrical potential difference (voltage) V, and power P.

Table II

	Resistance	Current	Voltage	Power
	W	Amps	Volt	Watt
Series	3.00		8.21	22.47
A	4.00	2.74	10.95	29.96
Series	9.00		10.14	11.43
B	8.00	1.13	9.02	10.16
Parallel	A 7.00	2.74		52.43
C	B 17.00	1.13	19.16	21.59
Series	C 4.96		19.16	74.02
D	2.00	3.86	7.73	29.86
Parallel	D 6.96	3.86		103.88
E	7.00	3.84	26.89	103.26
	6.00	4.48		120.48
Series	E 2.21		26.89	327.62
F	1.00	12.19	12.19	148.49
	5.00		60.93	742.45
Total	8.21	12.19	100.00	1218.57

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