Egypt Secondary 3 Physics, Week 2: Series and parallel together: worked circuits

The emf, internal resistance and the closed circuit

Week 2 · Chapter 1 (B): the emf of a cell and its internal resistance, Ohm’s law for a closed circuit I = V_B/(R + r), terminal voltage V = V_B − Ir and the emf as the open-circuit voltage, worked circuits (series, parallel, series-parallel, internal resistance) · 3 lessons

Series and parallel together: worked circuits

Physics Sec 3 Term 1, Chapter 1: pp. 13–17

GoalSolve a circuit with series and parallel parts in the right order: combine, find the total current, then go back to the branches.

New words

series–parallel circuit · a circuit with some resistors in series and a group of others in parallelcombine · replace a group of resistors by its equivalent resistancebranch current · the current in one resistor of a parallel group, found from the voltage across the group

Explanation

Most circuits mix both kinds of connection, so work in a fixed order. Step 1: replace each parallel group by its equivalent resistance. Step 2: add all the series resistances, including the internal resistance of the source, to get the total resistance. Step 3: find the total current, I = VB ÷ (total resistance). Step 4: go back to the circuit and find the voltage across each part with V = IR. Step 5: find each branch current from the voltage across its group. Step 6: check that the branch currents add up to the total current and that the voltages add up to the emf. The total current passes through every series part, while the branch currents only share it.

Step	What you do	Why
1	combine every parallel group	it becomes one resistor
2	add the series resistances, r included	this is the total resistance
3	I = VB ÷ total resistance	the total current
4	V = IR for each part	the voltage across each part
5	branch current = group voltage ÷ R	how the current splits
6	check the sums	currents add, voltages add

Worked examples

A 24 V source with negligible internal resistance feeds a 12 Ω and a 4 Ω resistor in parallel, in series with a 5 Ω resistor. The pair gives (12 × 4) ÷ 16 = 3 Ω. The total is 3 + 5 = 8 Ω, so I = 24 ÷ 8 = 3 A. The pair has 3 × 3 = 9 V and the 5 Ω resistor has 3 × 5 = 15 V (9 + 15 = 24 V). The branch currents are 9 ÷ 12 = 0.75 A and 9 ÷ 4 = 2.25 A, and 0.75 + 2.25 = 3 A.

A cell with VB = 14 V and r = 1 Ω feeds a 10 Ω and a 40 Ω resistor in parallel, in series with a 5 Ω resistor. The pair gives 400 ÷ 50 = 8 Ω. The outside resistance is 8 + 5 = 13 Ω and the total is 14 Ω, so I = 14 ÷ 14 = 1 A and the terminal voltage is 14 − 1 × 1 = 13 V. The pair has 1 × 8 = 8 V and the 5 Ω resistor has 5 V (8 + 5 = 13 V). The branches carry 8 ÷ 10 = 0.8 A and 8 ÷ 40 = 0.2 A, which add up to 1 A.

Working backwards: a 30 V source with negligible internal resistance is joined across a 20 Ω resistor in parallel with an unknown resistor R, and the total current is 2.5 A. The 20 Ω branch carries 30 ÷ 20 = 1.5 A, so the other branch carries 2.5 − 1.5 = 1.0 A and R = 30 ÷ 1.0 = 30 Ω.

Class activity

Draw the circuit of Example 2 on paper and write beside every resistor its current and its voltage. Which resistor carries the largest current, and which resistors have the largest voltage across them? Explain in two sentences why these are not the same resistor. Paper only.

Practice

A 40 V source with negligible internal resistance feeds 20 Ω and 30 Ω in parallel, in series with 8 Ω. Find the total current, the voltage across the pair and the branch currents.

A cell with VB = 10 V and r = 0.5 Ω is joined to an outside resistance of 4.5 Ω. Find the current and the terminal voltage.

A 24 V source with negligible internal resistance feeds a 20 Ω resistor in series with a 30 Ω and a 60 Ω resistor in parallel. Find the branch currents.

A 10 Ω lamp and a 40 Ω lamp are in parallel on the same cell. Which carries the larger current, and what is the ratio?