Finding the "drop in potential" as current moves through a single resistor of a circuit with parallel resistors

I have four different resistors connected in parallel to a $$\24.0 \ \text{V}\$$ battery with internal resistance of $$\0.500 \ \Omega\$$. The resistor values are $$\10.0 \ \Omega\$$,$$\15.0 \ \Omega\$$, $$\30.0 \ \Omega\$$, and $$\45.0 \ \Omega\$$.

I calculated the total resistance of the circuit to be the resistance of the four resistors in parallel, $$\\dfrac{1}{R_{\text{parallel}}} = \dfrac{1}{10} + \dfrac{1}{15} + \dfrac{1}{30} + \dfrac{1}{45} = \dfrac{1}{4.5} \ \Rightarrow R_{\text{parallel}} = 4.50 \ \Omega\$$, plus the internal resistance of the battery, so that $$\R_{\text{total}} = 4.50 \ \Omega + 0.500 \ \Omega = 5.00 \ \Omega \$$.

The voltage available at the terminals of the battery is found as follows: $$\I_{\text{total}} = \dfrac{V}{R_{\text{total}}} = 24.0/5.00 = 4.80 \ \text{A} \$$ for the total current, and then $$\V_{\text{parallel}} = I_{\text{total}} R_{\text{parallel}} = 4.80 \times 4.50 = 21.6 \ \text{V} \$$.

I now want to find the "potential difference across the $$\10.0 \ \Omega\$$ resistor. I am told that this is $$\V_{10} = V_{\text{parallel}} = 21.6 \ \text{V}\$$. But we know that Ohm's law can be interpreted as being the "drop in potential" as current flows through a resistor, so why is the solution not just $$\V_{10} = I_{\text{total}} \times 10 \ \Omega = 4.80 \ \text{A} \times 10 \ \Omega \$$? What is the reasoning behind why $$\V_{10} = V_{\text{parallel}} = 21.6 \ \text{V}\$$?

• The thing about applying Ohm's Law is identifying the correct current and the correct voltage for any given resistor. You can't just use any old current value existing somewhere in the wider circuit, you must use the current flowing through that particular resistor, and the voltage across that particular resistor. Commented May 13, 2022 at 7:50
• @Neil_UK Thanks for the clarifying advice. Commented May 14, 2022 at 7:47