K323 / 2027
D.C. circuits overview

Topic 4 of 6

Solving a complete circuit

Reduce a valid group to find the source current, then work back through the original connections to find each component's voltage and current.

Series groups share a path current. Parallel branches share endpoint voltage. R = V/I must refer to the same component or the same complete group.

Keep the same external endpoints

An equivalent resistance replaces a group at its two external terminals. It gives the same total current for the same p.d. across those terminals. It does not mean every original component had that current or the whole group's voltage.

Reduce the network while preserving terminal pairs

R1 is between A and B. The two parallel branches are between B and C. Keep those names when replacing a group or recovering its voltage.

Start with the complete network

Start with the complete networkA 12.0 V ideal source connects A to C through the complete external network. R1 is 4.0 ohm between A and B. R2 is 6.0 ohm between B and C, and R3 is 12.0 ohm between the same B and C nodes. R1 carries the source current, 1.50 A. R2 carries 1.0 A and R3 carries 0.50 A. Each series group has 6.0 V across it.+-12.0 VABCR1: 4.0 ohmR2: 6.0 ohmR3: 12.0 ohm1.50 A1.0 A0.50 AV_AB = 6.0 V; V_BC = 6.0 V

First replace only the B/C parallel group. Its two branches do not each receive the full 12.0 V.

Replace the B/C group by 4.0 ohm

Replace the B/C group by 4.0 ohmThe original 6.0 ohm and 12.0 ohm parallel branches between B and C are replaced by one 4.0 ohm equivalent. R1 remains 4.0 ohm between A and B. A and C are still the external source terminals. The two groups are in series, giving 8.0 ohm in total and a source current of 1.50 A at 12.0 V.+-12.0 VR1: 4 ohmB/C: 4 ohm6.0 V6.0 VABCI = 1.50 ASwitch closed

The drops 6.0 V and 6.0 V add to 12.0 V. Current is 1.50 A in the whole unbranched path.

Reduce the complete A/C network to 8.0 ohm

Reduce the complete A/C network to 8.0 ohmThe complete network is replaced by an 8.0 ohm equivalent between A and C across the same 12.0 V ideal source. The source current is 1.50 A. B was an internal connection and is absent after this final replacement; no separate component voltage is implied by the single equivalent resistor.+-12.0 VAC8.0 ohm1.50 A

Work back from the source current to the voltage across each original group, then to the individual branch currents.

Open only the 12 ohm branch

Open only the 12 ohm branchThe 12.0 V source still supplies R1, 4.0 ohm, between A and B. The 6.0 ohm branch between B and C remains connected. Only the 12.0 ohm branch has an open switch. Source current is now 1.20 A, all of which passes through the remaining 6.0 ohm branch. The open branch has no steady current. V_AB is 4.8 V and V_BC is 7.2 V.+-12.0 VABCR1: 4.0 ohmR2: 6.0 ohmR3: 12.0 ohm1.20 A1.20 A0 AV_AB = 4.8 V; V_BC = 7.2 V

The total current falls, but the current through the 6.0 ohm branch rises. Its shared series resistor changes the branch voltage.

R1 = 4.0 Ω connects A/B; the 6.0 Ω and 12.0 Ω branches each connect B/C. Replace only the B/C group first. The final comparison opens just the 12.0 Ω branch while leaving the 4.0 Ω and 6.0 Ω path intact.

Reduce, then work back

12.0 V across 4.0 Ω plus a parallel group

Assume an ideal source, negligible wire resistance and fixed component resistances.

  1. Reduce B/C: the 6.0 Ω and 12.0 Ω parallel equivalent is (6.0 x 12.0)/(6.0 + 12.0) = 4.0 Ω.
  2. Find the whole resistance: 4.0 + 4.0 = 8.0 Ω across A/C.
  3. Find source current: 12.0/8.0 = 1.50 A. This is also the current through R1.
  4. Find group voltages: VAB = 1.50 x 4.0 = 6.0 V. Therefore VBC = 12.0 - 6.0 = 6.0 V.
  5. Recover branch currents: 6.0/6.0 = 1.0 A through the 6.0 Ω resistor; 6.0/12.0 = 0.50 A through the 12.0 Ω resistor.

Check the result: 1.0 + 0.50 = 1.50 A at B, and 6.0 + 6.0 = 12.0 V along either complete path.

The parallel group has only 6.0 V across it here. Using 12.0 V for either branch would ignore the voltage across R1. Dividing the group's 6.0 V by total current 1.50 A finds the group's 4.0 Ω equivalent, not the resistance of either individual branch.

A branch current can rise while total current falls

Now open only the 12.0 Ω branch. The remaining path is 4.0 Ω in series with 6.0 Ω, so its total resistance is 10.0 Ω.

The new source current is 12.0/10.0 = 1.20 A. All of it passes through the remaining 6.0 Ω branch. That branch current has risen from 1.0 A to 1.20 A, even though source current has fallen from 1.50 A to 1.20 A.

R1 now has a smaller drop, 1.20 x 4.0 = 4.8 V. This leaves 7.2 V across the 6.0 Ω resistor. The changed voltage share explains its larger current. It was never connected directly across the ideal source.

Optional check An ideal 12.0 V source supplies a 4.0 ohm resistor in series with parallel 6.0 ohm and 12.0 ohm branches. Initially the source current is 1.50 A and the 6.0 ohm branch current is 1.0 A. Only the 12.0 ohm branch is then opened. What happens?
An ideal 12.0 V source supplies a 4.0 ohm resistor in series with parallel 6.0 ohm and 12.0 ohm branches. Initially the source current is 1.50 A and the 6.0 ohm branch current is 1.0 A. Only the 12.0 ohm branch is then opened. What happens?

Test the rules with paired measurements

Use a suitable low-voltage cell or battery, switch and appropriately rated resistors. Disconnect the source while changing connections. Use components whose heating is small enough for the intended constant-resistance comparison.

Match a voltage reading to the correct current reading

These are expected readings for the supplied low-voltage models, not experimental records. Meters are ideal: negligible ammeter resistance and negligible voltmeter current.

Measure the complete series network

Measure the complete series networkThe source drives an ammeter in the main positive lead before node A. A 10.0 ohm resistor connects A to B and a 20.0 ohm resistor connects B to C. A voltmeter connects across A and C, the complete pair. With ideal meters its reading is 3.00 V and the series ammeter reads 0.100 A. The individual resistor drops are 1.00 V and 2.00 V. The source return contains a closed switch.+-3.00 VA0.100 A+-B10.0 ohm20.0 ohm1.00 V2.00 VVAC+-3.00 VSwitch closed

Whole-network resistance: 3.00/0.100 = 30.0 ohm. The voltmeter spans both resistors, so its 3.00 V is not either single resistor drop.

Measure the complete parallel network

Measure the complete parallel networkThe source drives an ammeter in the main positive lead before node A. A 10.0 ohm resistor and a 20.0 ohm resistor are in separate branches from A to C. A voltmeter connects across A and C, the complete network. With ideal meters and 3.00 V across A/C, the branch currents are 0.300 A and 0.150 A and the main ammeter reads 0.450 A. The source return contains a closed switch.+-3.00 VA0.450 A+-10.0 ohm20.0 ohm0.300 A0.150 AVAC+-3.00 VSwitch closed

Whole-network resistance: 3.00/0.450 = 6.67 ohm. The main ammeter reads the sum of the two branch currents.

These separate measurement examples use 10.0 Ω and 20.0 Ω with a supplied terminal p.d. of 3.00 V. The main ammeter measures total current; the voltmeter across A/C measures the complete network's p.d. Component measurements need that component's own current and endpoints.

For the supplied 3.00 V model:

  • Series: total current is 0.100 A. The 10.0 Ω resistor has 1.00 V across it and the 20.0 Ω resistor 2.00 V. Whole-network resistance is 3.00/0.100 = 30.0 Ω.
  • Parallel: currents are 0.300 A through 10.0 Ω and 0.150 A through 20.0 Ω. Total current is 0.450 A, giving whole-network resistance 3.00/0.450 = 6.67 Ω.

These are expected values for the stated model. Record the actual current and terminal p.d. in an investigation. A real battery need not maintain exactly the same terminal voltage when the circuit changes.

  1. Choose the measurement. Put an ammeter in a main lead for total current, or in one selected branch for its current. Put the voltmeter across the exact component or group being investigated.
  2. Select the ranges and polarity. Use a suitable d.c. range. Conventional current should enter the ammeter's positive terminal; the voltmeter's positive lead goes to the higher-potential point. Check the zero and read an analogue scale without parallax. A nominal 3 V battery can exceed a 3 V range; establish a suitable range before choosing a finer one.
  3. Record positions with readings. Label the component or node pair and include units. Compare main current with the sum of branch currents, or complete p.d. with the series drops.
  4. Determine resistance from a matched pair. For one component, divide its p.d. by its current. For the whole network, divide the p.d. across A/C by total current entering it.

When moving one meter between positions, readings are taken at different times. Source drift and resistor heating can affect their comparison. Check the supply and allow conditions to settle; do not assume every discrepancy proves a failure of the circuit rule.

Loose contacts or contact resistance can change readings. Meter resistance, finite resolution and changes in temperature are separate limitations. Repeat readings to assess scatter, but correct a wrong connection or range directly. Finer resolution does not by itself guarantee better accuracy.

Brightness is not a calibrated current reading. A lamp's resistance changes as its filament heats. Do not treat a fixed-resistor calculation as an exact prediction for any lamp.