Question 18
A heater for a small greenhouse has three heating elements connected to a 12 V battery, as shown.
Calculate the total resistance of the circuit.
[2 marks] b)Calculate the power delivered by the battery.
[2 marks]Collapse the parallel pair, add the series element, then follow the current
The circuit is the printed one. Watch $R_1$ and $R_2$ become one 7.14 Ω resistance, then $R_3$ join it in series. Then watch 0.542 A leave the battery and split at the junction.
The dashes move at a speed proportional to the current in each wire, so they run 2.5 times as fast through $R_1$ as through $R_2$. The two branch currents add back to 0.542 A at the top junction.
Change the elements, then try the slips that cost marks
Move the sliders and the working updates live. The orange halo on each wire grows with the current in it. The question is $R_1=10\ \Omega$, $R_2=25\ \Omega$ and $R_3=15\ \Omega$. Then pick a common error to see the number it gives.
$R_1$ and $R_2$ join the same two junction dots, so they are in parallel, not in series. Adding all three makes $R_T$ too big, and it does not earn the second mark in a).
$R_3$ sits in the single wire from the battery, so every bit of current passes through it. It is in series with the pair, not in parallel with it. This $R_T$ does not earn the second mark in a).
The sum of reciprocals is $\dfrac{1}{R_{12}}$, in reciprocal ohms. Take the reciprocal to get $R_{12}$ before you add $R_3$. Check the size: a parallel pair always comes out between half the smaller element and the smaller element.
The 12 V is across the whole circuit, not across one element. The battery delivers power to all three, so use $R_T$. The working above shows how much $R_3$ really gets.
Pick an error to see the wrong working next to the correct numbers.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Combine the parallel pair first. Start from the parallel form in the formula book.
$$\begin{aligned}\frac{1}{R_T}&=\frac{1}{R_1}+\frac{1}{R_2}+\dots\\\frac{1}{R_{12}}&=\frac{1}{10}+\frac{1}{25}=0.14\\R_{12}&=\frac{1}{0.14}=7.143\ \Omega\end{aligned}$$
$R_1$ and $R_2$ are joined across the same two nodes, so they combine with the reciprocal form first.
Marker: provides correct formula and substituted values for the resistance of the parallel resistors
Add $R_3$ to the pair. Start from the series form in the formula book.
$$\begin{aligned}R_T&=R_1+R_2+R_3+\dots\\R_T&=R_{12}+R_3=7.143+15\\&=22.143\approx22.14\ \Omega\end{aligned}$$
All the current passes through $R_3$ and then the pair, so $R_3$ adds directly to $R_{12}$.
Allow FT marks for errors in prior working.
Accept total resistance range from 22.1–22.15 Ω inclusive.
Do not award the second mark for $R_T=50\ \Omega$ (all three resistors treated as in series) or $R_T=4.84\ \Omega$ (all three treated as in parallel).
Marker: correctly calculates the total resistance of the circuit
Find the current the battery supplies. Start from $V=IR$ and use the total resistance from a).
$$\begin{aligned}V&=IR\\I&=\frac{V}{R_T}=\frac{12}{22.14}=0.542\text{ A}\end{aligned}$$
The battery current is set by the full 12 V across the total resistance.
Marker: provides correct formula and substituted values for the current supplied by the battery
Multiply the battery voltage by the current it supplies. Start from $P=VI$.
$$\begin{aligned}P&=VI=12\times0.542\\&=6.504\text{ W}\approx6.50\text{ W}\end{aligned}$$
The power delivered by the battery is its voltage times the current it supplies. Write the unit, W, or this mark is lost.
Allow FT marks for errors in prior working.
Accept power range from 6.48–6.52 W inclusive.
Marker: correctly calculates the power delivered by the battery, with unit
Part a) is the same in both methods. Combine the parallel pair, then add $R_3$.
$$\begin{aligned}\frac{1}{R_{12}}&=\frac{1}{10}+\frac{1}{25}=0.14\\R_{12}&=7.143\ \Omega\\R_T&=R_{12}+R_3=7.143+15\approx22.14\ \Omega\end{aligned}$$
Marker: provides correct formula and substituted values for the resistance of the parallel resistors
Marker: correctly calculates the total resistance of the circuit
Start from $P=VI$, and write the current as $I=\dfrac{V}{R_T}$ from $V=IR$. Substitute it in.
$$\begin{aligned}P&=VI\quad\text{with}\quad I=\frac{V}{R_T}\\P&=V\times\frac{V}{R_T}=\frac{V^2}{R_T}=\frac{12^2}{22.14}\end{aligned}$$
The current formula, $I=\dfrac{V}{R_T}$ with 12 V and 22.14 Ω, is still on the page. It is folded into the power line instead of being worked out on its own line.
Method 2: $P=VI$ with $I=\dfrac{V}{R_T}$, so $P=\dfrac{12^2}{22.14}=6.50$ W.
Marker: provides correct formula and substituted values for the current supplied by the battery
Square the voltage first, then divide by the total resistance.
$$\begin{aligned}P&=\frac{144}{22.14}\\&=6.504\text{ W}\approx6.50\text{ W}\end{aligned}$$
This route never rounds the current, so it lands on the same 6.50 W with one less rounding step. The $R$ in $\dfrac{V^2}{R}$ must be $R_T$, because the 12 V is across the whole circuit.
Allow FT marks for errors in prior working.
Accept power range from 6.48–6.52 W inclusive.
Marker: correctly calculates the power delivered by the battery, with unit
Each phrase of the question gave you something
If I had to guess where you might lose marks…
- Getting the series and parallel parts wrong. You add all three resistors in series and get $R_T=50\ \Omega$, or you combine all three in parallel and get $R_T=4.84\ \Omega$. Neither earns Mark 2.
- Forgetting the reciprocal. You stop at $\dfrac{1}{R_{12}}=0.14$ and add it to $R_3$ to get $R_T=15.14\ \Omega$, instead of taking the reciprocal first to get $R_{12}=7.143\ \Omega$.
- Using one element instead of the whole circuit. You use $P=\dfrac{V^2}{R}$ with one element instead of $R_T$, for example $\dfrac{12^2}{15}=9.6$ W. The battery delivers power to the whole circuit, so use $R_T$.
- Leaving the unit off the power. You write 6.50 with no W, and that loses Mark 4.
New for 2026: QCAA has not examined this yet
No past QCAA question tests circuits, so compare the structure of the question instead.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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