Physics 9702 · Paper 2

Circuit Analysis

Collapse the network, then walk back

Combine the parallel pair that is furthest inside into one resistor.
Add anything in series with it, including r.
Total current from the supply: $I = E / R_{\text{total}}$ .
Walk back out: use $V = IR$ on each piece to find the p.d. or current the question actually asks for.

Worked example

Warm-up: one series resistor, one parallel pair

A battery of e.m.f. 4.5 V and negligible internal resistance (r is so small that you can ignore it) drives a 6.0 Ω resistor in series with a parallel pair: 12 Ω and 4.0 Ω. Find the current in the 4.0 Ω resistor.

Collapse the pair: $\dfrac{12 \times 4.0}{12 + 4.0} = 3.0 \; \Omega$ .
Total: $6.0 + 3.0 = 9.0 \; \Omega$ , so the main current is $I = 4.5/9.0 = 0.50$ A.
Walk back: the pair takes $V = 0.50 \times 3.0 = 1.5$ V, shared by both branches.
So the 4.0 Ω branch carries $I = 1.5/4.0 = 0.375 \approx 0.38$ A.

Answer

I \approx 0.38 \text{ A}

You can check the answer: the other branch carries $1.5/12 = 0.125$ A, and $0.375 + 0.125 = 0.50$ A, the main current. The first law confirms it.

Branch currents are often faster

Everything in parallel shares the same p.d.If you know that p.d., you can find each branch current directly and add them. No reciprocal formula needed.
Example: a 12 V supply with negligible internal resistance across 500 Ω and 1000 Ω in parallel. Branch currents: $12/500 = 0.024$ A and $12/1000 = 0.012$ A. Total drawn: $0.036$ A.
The same-p.d. rule also lets you ignore whole branches that the question does not ask about.

The full exam chain

Worked example

From terminal p.d. to an unknown resistor

Two resistors, 1.0 Ω and R, are in series with a cell of e.m.f. 1.50 V and internal resistance 0.28 Ω. The terminal p.d. across the cell is 1.36 V. Show that the current is 0.50 A, then find R.

Lost volts first: $E - V = 1.50 - 1.36 = 0.14$ V. All of it is across r.
$I = \dfrac{0.14}{0.28} = 0.50$ A, as required.
The external pair receives the terminal p.d., so together: $R_{\text{ext}} = \dfrac{1.36}{0.50} = 2.7 \; \Omega$ .
$R = 2.7 - 1.0 = 1.7 \; \Omega$ .

Answer

I = 0.50 \text{ A}, \; R = 1.7 \; \Omega

(9702/21/O/N/24 Q7)

Recognise the pattern: given E, r and the terminal p.d., the first step is always lost volts = E − V, and lost volts over r gives the current. Everything else follows with $V = IR$ .

Two loops need simultaneous equations

Some circuits cannot be collapsed, because two sources push currents through a shared branch. Then you use both laws directly:

Mark a current arrow on every branch. Guessed directions are fine.
First law at a junction: one equation linking the currents.
Second law around loops, one loop at a time, until you have as many equations as unknown currents.
Solve them as simultaneous equations.

Three branches, so three currents. Mark an arrow on every branch first. A wrong guess is fine: that current just comes out negative.

The circuit above, solved start to finish

First law at P: $I_1 + I_2 = I_3$ . Bottom loop (2.0 V cell and the 30 Ω): $2.0 = 30 I_3$ , so $I_3 = 0.067$ A. Top loop (6.0 V cell, the 10 Ω and the 30 Ω): $6.0 = 30 I_3 + 10 I_1 = 2.0 + 10 I_1$ , so $I_1 = 0.40$ A. Then $I_2 = I_3 - I_1 = -0.33$ A.

The outer loop would give a third equation, but it is just the top equation minus the bottom one: no new information. Two loops plus one junction is exactly enough for three currents.

A negative answer for a current means one thing only: you guessed its arrow backwards. Keep the size and reverse the direction.

PreviousSeries & Parallel Resistors NextEffect-of-change Questions

Circuit Analysis

Collapse the network, then walk back

Combine the parallel pair that is furthest inside into one resistor.
Add anything in series with it, including r.
Total current from the supply: $I = E / R_{\text{total}}$ .
Walk back out: use $V = IR$ on each piece to find the p.d. or current the question actually asks for.

Worked example

Warm-up: one series resistor, one parallel pair

Collapse the pair: $\dfrac{12 \times 4.0}{12 + 4.0} = 3.0 \; \Omega$ .
Total: $6.0 + 3.0 = 9.0 \; \Omega$ , so the main current is $I = 4.5/9.0 = 0.50$ A.
Walk back: the pair takes $V = 0.50 \times 3.0 = 1.5$ V, shared by both branches.
So the 4.0 Ω branch carries $I = 1.5/4.0 = 0.375 \approx 0.38$ A.

Answer

I \approx 0.38 \text{ A}

You can check the answer: the other branch carries $1.5/12 = 0.125$ A, and $0.375 + 0.125 = 0.50$ A, the main current. The first law confirms it.

Branch currents are often faster

Everything in parallel shares the same p.d.If you know that p.d., you can find each branch current directly and add them. No reciprocal formula needed.
Example: a 12 V supply with negligible internal resistance across 500 Ω and 1000 Ω in parallel. Branch currents: $12/500 = 0.024$ A and $12/1000 = 0.012$ A. Total drawn: $0.036$ A.
The same-p.d. rule also lets you ignore whole branches that the question does not ask about.

The full exam chain

Worked example

From terminal p.d. to an unknown resistor

Two resistors, 1.0 Ω and R, are in series with a cell of e.m.f. 1.50 V and internal resistance 0.28 Ω. The terminal p.d. across the cell is 1.36 V. Show that the current is 0.50 A, then find R.

Lost volts first: $E - V = 1.50 - 1.36 = 0.14$ V. All of it is across r.
$I = \dfrac{0.14}{0.28} = 0.50$ A, as required.
The external pair receives the terminal p.d., so together: $R_{\text{ext}} = \dfrac{1.36}{0.50} = 2.7 \; \Omega$ .
$R = 2.7 - 1.0 = 1.7 \; \Omega$ .

Answer

I = 0.50 \text{ A}, \; R = 1.7 \; \Omega

(9702/21/O/N/24 Q7)

Recognise the pattern: given E, r and the terminal p.d., the first step is always lost volts = E − V, and lost volts over r gives the current. Everything else follows with $V = IR$ .

Two loops need simultaneous equations

Some circuits cannot be collapsed, because two sources push currents through a shared branch. Then you use both laws directly:

Mark a current arrow on every branch. Guessed directions are fine.
First law at a junction: one equation linking the currents.
Second law around loops, one loop at a time, until you have as many equations as unknown currents.
Solve them as simultaneous equations.

Three branches, so three currents. Mark an arrow on every branch first. A wrong guess is fine: that current just comes out negative.

The circuit above, solved start to finish

The outer loop would give a third equation, but it is just the top equation minus the bottom one: no new information. Two loops plus one junction is exactly enough for three currents.

A negative answer for a current means one thing only: you guessed its arrow backwards. Keep the size and reverse the direction.