The voltage of a circuit is determined by the potential difference of the source. If I take a $4.5 \ \text{V}$ battery and short-circuit it, then theoretically (neglecting the internal resistance of the source and the reaction speed of the chemical processes), all releasable electrons flow from the source to the sink at once and at a voltage of $4.5 \ \text{V}$.

If I now connect the current source to a conductor, I also connect a resistor between the potential difference of the battery. The potential difference of the battery remains the same, as this is caused by its chemical process. What changes is the conductivity for the number of electrons.

If you now measure the potential difference to one pole of the source along your wire at ten equally distant points, you will find that the voltage along the wire changes by $0.45 \ \text{V}$ from measurement to measurement. If you use a real resistor, this, of course, radically changes the game. The ohmic resistance of the lead wire is usually not considered relevant, and the entire potential difference is attributed to the resistor.

My question is, why is it that the sum of voltages across both resistors is the original voltage?

If you measure the potential difference along your resistor at every tenth of a distance, you will also get a voltage difference of $0.45 \ \text{V}$, as in the above example with the wire (which is now neglected). If you divide your resistor into two equal resistors, each will generate a voltage difference of $2.25 \ \text{V}$.

The voltage of a circuit is determined by the potential difference of the source. If I take a $4.5 \,\text{V}$ battery and short-circuit it, then theoretically (neglecting the internal resistance of the source and the reaction speed of the chemical processes), all releasable electrons flow from the source to the sink at once and at a voltage of $4.5 \,\text{V}$.

If I now connect the current source to a conductor, I also connect a resistor between the potential difference of the battery. The potential difference of the battery remains the same, as this is caused by its chemical process. What changes is the conductivity for the number of electrons.

If you now measure the potential difference to one pole of the source along your wire at ten equally distant points, you will find that the voltage along the wire changes by $0.45 \,\text{V}$ from measurement to measurement. If you use a real resistor, this, of course, radically changes the game. The ohmic resistance of the lead wire is usually not considered relevant, and the entire potential difference is attributed to the resistor.

My question is, why is it that the sum of voltages across both resistors is the original voltage?

If you measure the potential difference along your resistor at every tenth of a distance, you will also get a voltage difference of $0.45 \,\text{V}$, as in the above example with the wire (which is now neglected). If you divide your resistor into two equal resistors, each will generate a voltage difference of $2.25 \,\text{V}$.

The voltage of a circuit is determined by the potential difference of the source. If I take a 4.5V battery and short-circuit it, then theoretically (neglecting the internal resistance of the source and the reaction speed of the chemical processes) all releasable electrons flow from the source to the sink at once, and that at a voltage of 4.5V.

If I now connect the current source to a conductor, I also connect a resistor between the potential difference of the battery. The potential difference of the battery remains the same, as this is caused by its chemical process. What changes is the conductivity for the number of electrons.

If you now measure the potential difference to one pole of the source along your wire at 10 equally distant points, you will find that the voltage along the wire changes by 0.45 V from measurement to measurement. If you use a real resistor, this of course radically changes the game. The ohmic resistance of the lead wire is usually not considered relevant and the entire potential difference is attributed to the resistor.

My question is, why is it that the sum of voltages across both resistors the original voltage?

If you measure the potential difference along your resistor at every tenth of a distance, you will also get a voltage difference of 0.45 V, as in the above example with the wire (which is now neglected). If you divide your resistor into two equal resistors, each will generate a voltage difference of 2.25 V.

The voltage of a circuit is determined by the potential difference of the source. If I take a $4.5 \ \text{V}$ battery and short-circuit it, then theoretically (neglecting the internal resistance of the source and the reaction speed of the chemical processes), all releasable electrons flow from the source to the sink at once and at a voltage of $4.5 \ \text{V}$.

If I now connect the current source to a conductor, I also connect a resistor between the potential difference of the battery. The potential difference of the battery remains the same, as this is caused by its chemical process. What changes is the conductivity for the number of electrons.

If you now measure the potential difference to one pole of the source along your wire at ten equally distant points, you will find that the voltage along the wire changes by $0.45 \ \text{V}$ from measurement to measurement. If you use a real resistor, this, of course, radically changes the game. The ohmic resistance of the lead wire is usually not considered relevant, and the entire potential difference is attributed to the resistor.

My question is, why is it that the sum of voltages across both resistors is the original voltage?

If you measure the potential difference along your resistor at every tenth of a distance, you will also get a voltage difference of $0.45 \ \text{V}$, as in the above example with the wire (which is now neglected). If you divide your resistor into two equal resistors, each will generate a voltage difference of $2.25 \ \text{V}$.