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The emf, internal resistance and the closed circuit

الأسبوع 2 · الفصل الأول (٢): القوة الدافعة الكهربية والمقاومة الداخلية وقانون أوم للدائرة المغلقة وفرق الجهد بين قطبي المصدر، وأمثلة محلولة على التوالي والتوازي ومقاومة المصدر الداخلية · 3 دروس
3

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.

StepWhat you doWhy
1combine every parallel groupit becomes one resistor
2add the series resistances, r includedthis is the total resistance
3I = VB ÷ total resistancethe total current
4V = IR for each partthe voltage across each part
5branch current = group voltage ÷ Rhow the current splits
6check the sumscurrents add, voltages add
Worked examples
12 Ω4 Ω24 V+−5 Ω
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
1
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.
2
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.
3
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.
4
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?