Question 7
A circuit containing a battery and two resistors is shown.
What is the total current supplied by the battery?
Not quite. You used $R_T = R_1 + R_2 = 10\ \Omega$, which is the series rule, giving $\dfrac{12}{10} = 1.2\text{ A}$.
Not quite. You found the current in the 4 Ω branch only: $\dfrac{12}{4} = 3\text{ A}$.
That is it. Step through the hints below to see why.
Not quite. You used $\dfrac{1}{R_T} = \dfrac{5}{12}$ as $R_T$ without inverting it: $12 \div \dfrac{5}{12} = 28.8\text{ A}$.
The printed circuit with current flowing. Orange dots: conventional current flowing out of the battery’s long (positive) plate and around the circuit, faster where the current is larger. Shaded orange: the branches the option you are testing counts. Dashed blue boxes: the two resistors added as if they were in series.
1.2 A is less than the 3 A the 4 Ω branch draws on its own, so it cannot be the total. The series rule pictures one path, but this circuit has two side by side.
3 A is only the current through the 4 Ω branch. The 6 Ω branch carries another 2 A, and the battery supplies both.
Each branch has the full 12 V across it: 3 A flows through 4 Ω and 2 A through 6 Ω. Together they make the 5 A the battery supplies.
28.8 A is far more than the 3 A and 2 A the branches draw. $\dfrac{5}{12}$ is $\dfrac{1}{R_T}$, so you must invert it to get $R_T = 2.4\ \Omega$.
Both resistors connect directly across the battery. Find the total resistance with the parallel rule.
$$\begin{aligned}\frac{1}{R_T}&=\frac{1}{R_1}+\frac{1}{R_2}\\&=\frac{1}{4}+\frac{1}{6}=\frac{5}{12}\\R_T&=\frac{12}{5}=2.4\ \Omega\end{aligned}$$
Both resistors connect directly across the battery, so they are in parallel.
Use $V=IR$ with the total resistance.
$$\begin{aligned}V&=IR\\I&=\frac{V}{R_T}=\frac{12}{2.4}=5\text{ A}\end{aligned}$$
Check: each branch has 12 V across it, so $\dfrac{12}{4}=3\text{ A}$ and $\dfrac{12}{6}=2\text{ A}$, and $3+2=5\text{ A}$. This is option (C).
The resistors are in parallel, so $R_T = 2.4\ \Omega$, and the battery supplies $I=\dfrac{V}{R_T}=\dfrac{12}{2.4}=5\text{ A}$.
New for 2026: QCAA has not examined this yet
No past QCAA question tests circuits, so there is no past question to compare this one with.
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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