3.7 A little more on potential difference - effect of two resistors in series - explaining and verifying by calculation, how the potential difference is split between resistances in series

The circuit 41 shows two resistors wired in series, so how is the total p.d. divided or shared between the two resistance wired in series.

This circuit can be used to investigate and confirm the correctness of the following theoretical calculations.

On the right is shown what happens to the p.d. going clockwise around the circuit (direction of convention current).

The potential store of the battery raises the potential difference of the charge to 12.0 V.

The total resistance = 10.0 + 5.0 = 15.0 Ω (you can add resistors up if wired in series)

Therefore the current flowing round all of this series circuit = I = V / R_total = 12.0 / 15.0 = 0.80 A

As the charge passes through the 1st resistor R₁, it loses energy and the p.d. falls by 8 V to a p.d. of 4 V.

Calculation to confirm this:  V₁ = I x R₁ = 0.80 x 10.0 = 8.0 V

As the charge passes through the 2nd resistor R₁, it loses energy again and the p.d. falls by 4 V to a p.d. of 0 V.

Calculation to confirm this:  V₂ = I x R₂ = 0.80 x 5.0 = 4.0 V

As long as there is a complete circuit, the process repeats itself through any number of resistors.

What is clear is the total p.d. for the circuit is split into a p.d. ratio identical to the resistor ratio.

V₁ : V₂ is the same ratio as R₁ : R₂

You can also say that since E = QV, twice as much energy is released by resistor R₁ (p.d. 8 V) than R₂ (p.d. 4 V) for the same total current, the same total charge transferred, but the lower the p.d. the less potential energy carried by the charge.

Be able to explain and verifying by calculation (and experiment), how potential difference is split shared divided between resistances in series.

Be able to describe the effect of two resistors in series on the p.d. across these resistors wired in series.

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