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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?

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2 Answers 2

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21.6V is the voltage across your external resistance, after the internal resistance of the battery. If you are taking the - terminal of the battery to be your ground (0V), the voltage at the + terminal will be 21.6V.

schematic

simulate this circuit – Schematic created using CircuitLab

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The voltage between the terminals of the [battery plus internal resistance] is 21.6 V.

If the negative terminal of the [battery plus internal resistance] is Grounded, it is considered to be Zero Volts, so the positive terminal of the [battery plus internal resistance] will be at 21.6 V.

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