# Finding the voltage available at the terminals of the battery in 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 \$$.

I am now tasked with finding the "voltage available at the terminals of the battery." The solution is given as $$\I_{\text{total}} = \dfrac{V}{R_{\text{total}}} = 24.0/5.00 = 4.80 \ \text{A} \$$ for the total current, and then $$\V_{\text{parallel}} = I_{\text{total}} R_{\text{parallel}} = 4.80 \times 4.50 = 21.6 \ \text{V} \$$. So does this mean that the voltage at the "end-point" (the negative terminal) of the battery is $$\21.6 \ \text{V}\$$? And would the voltage at the beginning (the positive terminal) just be $$\24 \ \text{V}\$$? What is the reasoning behind this solution? 