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To find the potential difference between points A and C, we first analyze the circuit. The circuit consists of two parallel branches connected to a voltage source. One branch contains resistors of $6\,\Omega$ and $15\,\Omega$ in series, and the other branch contains resistors of $10\,\Omega$ and $30\,\Omega$ in series.

Let the total resistance of the first branch be $R_1$ and the second branch be $R_2$:

$$ \begin{aligned} R_1 &= 6\,\Omega + 15\,\Omega = 21\,\Omega \\ R_2 &= 10\,\Omega + 30\,\Omega = 40\,\Omega \end{aligned} $$

Since these two branches are connected in parallel across the $12\,\text{V}$ source, the voltage across each branch is $12\,\text{V}$.

We need to find the potential difference between points A and C. Point A is at the start of the second branch, and point C is between the $10\,\Omega$ and $30\,\Omega$ resistors in that same branch. The current $I_2$ flowing through the second branch is:

$$ \begin{aligned} I_2 &= \frac{V}{R_2} \\ &= \frac{12\,\text{V}}{40\,\Omega} \\ &= 0.3\,\text{A} \end{aligned} $$

The potential difference between points A and C ($V_{AC}$) is the voltage drop across the $10\,\Omega$ resistor:

$$ \begin{aligned} V_{AC} &= I_2 \times 10\,\Omega \\ &= 0.3\,\text{A} \times 10\,\Omega \\ &= 3\,\text{V} \end{aligned} $$

Therefore, the potential difference between points A and C is $3\,\text{V}$.