Physics 9702 · Paper 2

Series & Parallel Resistors

The two resistor-combination formulas both come straight from Kirchhoff's laws, and the exam asks for these short derivations.

Series: one path

Series: one path, same current, p.d.s add up. Parallel: two paths, same p.d., currents split.

One path, so the current I is the same through both resistors (first law: no junction, nowhere to split).
The p.d.s add (second law): $V = V_1 + V_2$ , so $IR = IR_1 + IR_2$ .
Cancel the common factor I and the formula appears.

R = R_1 + R_2 + \dots

Symbols

$R$ = combined resistance in series (\Omega)
$R_1, R_2$ = individual resistances (\Omega)

Worked example

Warm-up: three in a row

Find the combined resistance of 5 Ω, 5 Ω and 10 Ω in series.

$R = 5 + 5 + 10 = 20 \; \Omega$ .

Answer

R = 20 \; \Omega

Parallel: two paths

Two paths between the same two points, so the p.d. V is the same across both (second law: a loop through the two branches contains no e.m.f.).
The currents add (first law): $I = I_1 + I_2$ , so $\dfrac{V}{R} = \dfrac{V}{R_1} + \dfrac{V}{R_2}$ .
Cancel the common factor V and the formula appears.

\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots

Symbols

$R$ = combined resistance in parallel (\Omega)
$R_1, R_2$ = individual resistances (\Omega)

Worked example

Warm-up: two equal resistors

Find the combined resistance of two 10 Ω resistors in parallel.

$\dfrac{1}{R} = \dfrac{1}{10} + \dfrac{1}{10} = \dfrac{2}{10}$ .
Turn $1/R$ back into R: $R = 5 \; \Omega$ .

Answer

R = 5 \; \Omega

Common mistake

Stopping at

\dfrac{1}{R} = 0.2

and writing R = 0.2 Ω. The reciprocal formula gives you 1/R, so the last step is always to turn 1/R back into R: R = 5 Ω. This one mistake loses the mark more often than any algebra error in the topic.

Both derivations are the two laws in one line each. Series: same I, p.d.s add. Parallel: same V, currents add. Write those words and you get both marks.

Quick checks that find wrong answers

A parallel combination is always smaller than the smallest branch. An extra branch is an extra path for current, so the combination conducts better. If your parallel answer is bigger than any branch, it is wrong.
For exactly two resistors: $R = \dfrac{R_1 R_2}{R_1 + R_2}$ , product over sum. Faster, and no reciprocal to forget.
Equal resistors in parallel: two of them give half the value of one, three give a third.

Worked example

A mixed combination

A 100 Ω and a 200 Ω resistor are in series, and the pair is in parallel with another 200 Ω resistor. Find the total resistance.

Series pair first: $100 + 200 = 300 \; \Omega$ .
Then product over sum: $R = \dfrac{300 \times 200}{300 + 200} = 120 \; \Omega$ .
Check: 120 Ω is smaller than the smallest branch (200 Ω), as a parallel answer must be.

Answer

R = 120 \; \Omega

Your turn— tap to reveal the worked answer (9702/11/M/J/25 Q36)

Two resistors R1 and R2 are connected in parallel between X and Y. Which expression gives the combined resistance?

Answer: $\dfrac{R_1 R_2}{R_1 + R_2}$ . The option $\dfrac{R_1 + R_2}{R_1 R_2}$ is the same fraction upside down, put there for people who do not remember the formula exactly. (9702/11/M/J/25 Q36)

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Series & Parallel Resistors

The two resistor-combination formulas both come straight from Kirchhoff's laws, and the exam asks for these short derivations.

Series: one path

Series: one path, same current, p.d.s add up. Parallel: two paths, same p.d., currents split.

One path, so the current I is the same through both resistors (first law: no junction, nowhere to split).
The p.d.s add (second law): $V = V_1 + V_2$ , so $IR = IR_1 + IR_2$ .
Cancel the common factor I and the formula appears.

R = R_1 + R_2 + \dots

Symbols

$R$ = combined resistance in series (\Omega)
$R_1, R_2$ = individual resistances (\Omega)

Worked example

Warm-up: three in a row

Find the combined resistance of 5 Ω, 5 Ω and 10 Ω in series.

$R = 5 + 5 + 10 = 20 \; \Omega$ .

Answer

R = 20 \; \Omega

Parallel: two paths

Two paths between the same two points, so the p.d. V is the same across both (second law: a loop through the two branches contains no e.m.f.).
The currents add (first law): $I = I_1 + I_2$ , so $\dfrac{V}{R} = \dfrac{V}{R_1} + \dfrac{V}{R_2}$ .
Cancel the common factor V and the formula appears.

\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots

Symbols

$R$ = combined resistance in parallel (\Omega)
$R_1, R_2$ = individual resistances (\Omega)

Worked example

Warm-up: two equal resistors

Find the combined resistance of two 10 Ω resistors in parallel.

$\dfrac{1}{R} = \dfrac{1}{10} + \dfrac{1}{10} = \dfrac{2}{10}$ .
Turn $1/R$ back into R: $R = 5 \; \Omega$ .

Answer

R = 5 \; \Omega

Common mistake

Stopping at

\dfrac{1}{R} = 0.2

Both derivations are the two laws in one line each. Series: same I, p.d.s add. Parallel: same V, currents add. Write those words and you get both marks.

Quick checks that find wrong answers

A parallel combination is always smaller than the smallest branch. An extra branch is an extra path for current, so the combination conducts better. If your parallel answer is bigger than any branch, it is wrong.
For exactly two resistors: $R = \dfrac{R_1 R_2}{R_1 + R_2}$ , product over sum. Faster, and no reciprocal to forget.
Equal resistors in parallel: two of them give half the value of one, three give a third.

Worked example

A mixed combination

A 100 Ω and a 200 Ω resistor are in series, and the pair is in parallel with another 200 Ω resistor. Find the total resistance.

Series pair first: $100 + 200 = 300 \; \Omega$ .
Then product over sum: $R = \dfrac{300 \times 200}{300 + 200} = 120 \; \Omega$ .
Check: 120 Ω is smaller than the smallest branch (200 Ω), as a parallel answer must be.

Answer

R = 120 \; \Omega

Your turn— tap to reveal the worked answer (9702/11/M/J/25 Q36)

Two resistors R1 and R2 are connected in parallel between X and Y. Which expression gives the combined resistance?

Answer: $\dfrac{R_1 R_2}{R_1 + R_2}$ . The option $\dfrac{R_1 + R_2}{R_1 R_2}$ is the same fraction upside down, put there for people who do not remember the formula exactly. (9702/11/M/J/25 Q36)