Topics

Combination circuits

Covers AS91173

The circuit this standard examines

AS91173 limits DC electricity to parallel circuits with resistive components in series with the source.
That means the standard exam circuit is a parallel group connected in series with one or more other resistors and the supply.
Everything is solved with the rules you already have — the only new skill is applying them in the right order.

The method

Work inward to find the total resistance, then outward to find everything else:

Simplify the parallel group into a single equivalent resistance, using $\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ .
Add that equivalent resistance to the series resistances: $R_T = R_{\text{series}} + R_P$ .
Find the total current from the supply: $I_T = \dfrac{V_T}{R_T}$ .
Find the voltage across the series resistor: $V = I_TR_{\text{series}}$ . The total current flows through it.
Find the voltage across the parallel group: either $V_P = I_TR_P$ , or subtract the series voltage from the supply voltage.
Find each branch current using that parallel-group voltage: $I_1 = \dfrac{V_P}{R_1}$ , and so on.
Check: the branch currents must add to the total current, and the voltages must add to the supply voltage.

The two traps

The full supply voltage is not across the parallel group. Some of it is dropped across the series resistor first. Using the supply voltage to find the branch currents is the single most common error in this topic.
The total current is not the branch current. The total current flows through the series resistor; it splits in the parallel group.

What happens when the circuit changes

Questions frequently ask you to predict the effect of a change. Reason it through in this order — total resistance, then total current, then voltages, then branch currents:

Adding another branch to the parallel group:
- $R_P$ falls, so $R_T$ falls, so $I_T$ increases,
- the voltage across the series resistor increases (bigger current through the same resistance),
- so the voltage across the parallel group decreases,
- so the current in each existing branch falls slightly — even though nothing about those branches changed.
Removing a branch does the opposite.
A branch breaking (a bulb blowing) removes that path: $R_P$ rises, $I_T$ falls, and the remaining branches get more voltage and so more current — they get brighter.
This chain of reasoning is where the Excellence marks are. Work through it link by link rather than guessing the end result.

Worked ExampleA parallel pair in series with a resistor

A $12$ V battery is connected in series with a $4.0$ Ω resistor, which is then connected to a parallel combination of a $12$ Ω and a $6.0$ Ω resistor. Find the total resistance, the total current, the voltage across each part, and the current in each branch.

Step 1 — Simplify the parallel group

$\frac{1}{R_P} = \frac{1}{12} + \frac{1}{6.0} = 0.08333 + 0.16667 = 0.25000$ $R_P = 4.0 \; \Omega$

Step 2 — Total resistance

$R_T = R_{\text{series}} + R_P = 4.0 + 4.0 = 8.0 \; \Omega$

Step 3 — Total current

$I_T = \frac{V_T}{R_T} = \frac{12}{8.0} = 1.5 \text{ A}$

Step 4 — Voltage across each part

Across the series resistor (the total current flows through it):

$V_{\text{series}} = I_TR_{\text{series}} = (1.5)(4.0) = 6.0 \text{ V}$

Across the parallel group:

$V_P = 12 - 6.0 = 6.0 \text{ V}$

Step 5 — Branch currents, using $V_P$ (not the supply voltage)

$I_1 = \frac{V_P}{R_1} = \frac{6.0}{12} = 0.50 \text{ A}$ $I_2 = \frac{V_P}{R_2} = \frac{6.0}{6.0} = 1.0 \text{ A}$

Step 6 — Check

$I_1 + I_2 = 0.50 + 1.0 = 1.5 \text{ A} = I_T \;\checkmark$

$R_T = 8.0 \; \Omega, \; I_T = 1.5 \text{ A}, \; V_{\text{series}} = V_P = 6.0 \text{ V}, \; I_1 = 0.50 \text{ A}, \; I_2 = 1.0 \text{ A}$

Worked ExamplePredicting the effect of a change

In the circuit above, the $6.0$ Ω branch is disconnected. Predict and calculate what happens to the total current and to the current in the $12$ Ω branch.

Step 1 — Predict, using the chain of reasoning

Removing a parallel branch raises $R_P$ , which raises $R_T$ , which lowers $I_T$ . A lower total current means less voltage across the series resistor, and therefore more voltage across the remaining branch — so the current in the $12$ Ω branch should increase.

Step 2 — New total resistance

With only the $12$ Ω branch left, the parallel group is simply $12$ Ω:

$R_T = 4.0 + 12 = 16 \; \Omega$

Step 3 — New total current

$I_T = \frac{12}{16} = 0.75 \text{ A}$

Down from $1.5$ A, as predicted.

Step 4 — New voltage across the $12$ Ω resistor

$V_{\text{series}} = (0.75)(4.0) = 3.0 \text{ V}$ $V_P = 12 - 3.0 = 9.0 \text{ V}$

Step 5 — Current in the $12$ Ω branch

$I_1 = \frac{9.0}{12} = 0.75 \text{ A}$

Up from $0.50$ A — and equal to $I_T$ , as it must be now that there is only one path.

$I_T: 1.5 \to 0.75 \text{ A}, \qquad I_{12\Omega}: 0.50 \to 0.75 \text{ A}$

Combination circuits

Covers AS91173

The circuit this standard examines

AS91173 limits DC electricity to parallel circuits with resistive components in series with the source.
That means the standard exam circuit is a parallel group connected in series with one or more other resistors and the supply.
Everything is solved with the rules you already have — the only new skill is applying them in the right order.

The method

Work inward to find the total resistance, then outward to find everything else:

Simplify the parallel group into a single equivalent resistance, using $\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ .
Add that equivalent resistance to the series resistances: $R_T = R_{\text{series}} + R_P$ .
Find the total current from the supply: $I_T = \dfrac{V_T}{R_T}$ .
Find the voltage across the series resistor: $V = I_TR_{\text{series}}$ . The total current flows through it.
Find the voltage across the parallel group: either $V_P = I_TR_P$ , or subtract the series voltage from the supply voltage.
Find each branch current using that parallel-group voltage: $I_1 = \dfrac{V_P}{R_1}$ , and so on.
Check: the branch currents must add to the total current, and the voltages must add to the supply voltage.

The two traps

The full supply voltage is not across the parallel group. Some of it is dropped across the series resistor first. Using the supply voltage to find the branch currents is the single most common error in this topic.
The total current is not the branch current. The total current flows through the series resistor; it splits in the parallel group.

What happens when the circuit changes

Questions frequently ask you to predict the effect of a change. Reason it through in this order — total resistance, then total current, then voltages, then branch currents:

Adding another branch to the parallel group:
- $R_P$ falls, so $R_T$ falls, so $I_T$ increases,
- the voltage across the series resistor increases (bigger current through the same resistance),
- so the voltage across the parallel group decreases,
- so the current in each existing branch falls slightly — even though nothing about those branches changed.
Removing a branch does the opposite.
A branch breaking (a bulb blowing) removes that path: $R_P$ rises, $I_T$ falls, and the remaining branches get more voltage and so more current — they get brighter.
This chain of reasoning is where the Excellence marks are. Work through it link by link rather than guessing the end result.

Worked ExampleA parallel pair in series with a resistor

Step 1 — Simplify the parallel group

$\frac{1}{R_P} = \frac{1}{12} + \frac{1}{6.0} = 0.08333 + 0.16667 = 0.25000$ $R_P = 4.0 \; \Omega$

Step 2 — Total resistance

$R_T = R_{\text{series}} + R_P = 4.0 + 4.0 = 8.0 \; \Omega$

Step 3 — Total current

$I_T = \frac{V_T}{R_T} = \frac{12}{8.0} = 1.5 \text{ A}$

Step 4 — Voltage across each part

Across the series resistor (the total current flows through it):

$V_{\text{series}} = I_TR_{\text{series}} = (1.5)(4.0) = 6.0 \text{ V}$

Across the parallel group:

$V_P = 12 - 6.0 = 6.0 \text{ V}$

Step 5 — Branch currents, using $V_P$ (not the supply voltage)

$I_1 = \frac{V_P}{R_1} = \frac{6.0}{12} = 0.50 \text{ A}$ $I_2 = \frac{V_P}{R_2} = \frac{6.0}{6.0} = 1.0 \text{ A}$

Step 6 — Check

$I_1 + I_2 = 0.50 + 1.0 = 1.5 \text{ A} = I_T \;\checkmark$

$R_T = 8.0 \; \Omega, \; I_T = 1.5 \text{ A}, \; V_{\text{series}} = V_P = 6.0 \text{ V}, \; I_1 = 0.50 \text{ A}, \; I_2 = 1.0 \text{ A}$

Worked ExamplePredicting the effect of a change

In the circuit above, the $6.0$ Ω branch is disconnected. Predict and calculate what happens to the total current and to the current in the $12$ Ω branch.

Step 1 — Predict, using the chain of reasoning

Step 2 — New total resistance

With only the $12$ Ω branch left, the parallel group is simply $12$ Ω:

$R_T = 4.0 + 12 = 16 \; \Omega$

Step 3 — New total current

$I_T = \frac{12}{16} = 0.75 \text{ A}$

Down from $1.5$ A, as predicted.

Step 4 — New voltage across the $12$ Ω resistor

$V_{\text{series}} = (0.75)(4.0) = 3.0 \text{ V}$ $V_P = 12 - 3.0 = 9.0 \text{ V}$

Step 5 — Current in the $12$ Ω branch

$I_1 = \frac{9.0}{12} = 0.75 \text{ A}$

Up from $0.50$ A — and equal to $I_T$ , as it must be now that there is only one path.

$I_T: 1.5 \to 0.75 \text{ A}, \qquad I_{12\Omega}: 0.50 \to 0.75 \text{ A}$

Combination circuits

The circuit this standard examines

The method

The two traps

What happens when the circuit changes

Step 1 — Simplify the parallel group

Step 2 — Total resistance

Step 3 — Total current

Step 4 — Voltage across each part

Step 5 — Branch currents, using VPV_PVP​ (not the supply voltage)

Step 6 — Check

Step 1 — Predict, using the chain of reasoning

Step 2 — New total resistance

Step 3 — New total current

Step 4 — New voltage across the 121212 Ω resistor

Step 5 — Current in the 121212 Ω branch

Combination circuits

The circuit this standard examines

The method

The two traps

What happens when the circuit changes

Step 1 — Simplify the parallel group

Step 2 — Total resistance

Step 3 — Total current

Step 4 — Voltage across each part

Step 5 — Branch currents, using VPV_PVP​ (not the supply voltage)

Step 6 — Check

Step 1 — Predict, using the chain of reasoning

Step 2 — New total resistance

Step 3 — New total current

Step 4 — New voltage across the 121212 Ω resistor

Step 5 — Current in the 121212 Ω branch

Step 5 — Branch currents, using $V_P$ (not the supply voltage)

Step 4 — New voltage across the $12$ Ω resistor

Step 5 — Current in the $12$ Ω branch

Step 5 — Branch currents, using $V_P$ (not the supply voltage)

Step 4 — New voltage across the $12$ Ω resistor

Step 5 — Current in the $12$ Ω branch