42-44 - Potential Dividers and Sensors

42-44 - Potential Dividers and Sensors

Use a uniform wire and resistor ratios to select a potential difference, then predict how light and temperature sensors change a divider output.

Potential Along a Uniform Wire

Imagine a uniform resistance wire stretched between the two terminals of a low-voltage supply. The wire has the same material and cross-section all the way along, so each centimetre has the same resistance as any other centimetre.

When a steady current flows, the potential changes at a steady rate along the wire. If one end is defined as 0 V0\,\mathrm{V} and the other end is at the supply p.d. VsV_{\mathrm{s}}, then the potential at distance xx from the 0 V0\,\mathrm{V} end is:

Potential along a uniform wire

Vx=VsxLΔV=VsΔxLV_x = V_{\mathrm{s}} \frac{x}{L} \qquad \Delta V = V_{\mathrm{s}} \frac{\Delta x}{L}

VsV_{\mathrm{s}}: p.d. across the whole wire (V); xx: distance from the 0 V end (m); LL: total length (m). For a uniform wire carrying a steady current.

The fraction of the total length gives the fraction of the total p.d., and the p.d. between two points is found from the distance between them.

This works because the wire is uniform and current-carrying. Equal lengths of the wire have equal resistance, so equal lengths take equal shares of the p.d. It is not that current is being used up as it moves along the wire. The current is the same through every part of the series path; the potential changes because energy is transferred from each coulomb of charge to the wire.

Worked route:

A 1.20 m1.20\,\mathrm{m} uniform wire is connected across a 6.0 V6.0\,\mathrm{V} supply. The left end is at 0 V0\,\mathrm{V}. Determine the potential at a point 0.40 m0.40\,\mathrm{m} from the left end, and the p.d. between points 0.25 m0.25\,\mathrm{m} and 0.85 m0.85\,\mathrm{m} from the left end.

For the first point:

Vx=6.0×0.401.20=2.0 VV_x = 6.0 \times \frac{0.40}{1.20} = 2.0\,\mathrm{V}

For the p.d. between two points, use the separation along the wire:

Δx=0.85−0.25=0.60 m\Delta x = 0.85 - 0.25 = 0.60\,\mathrm{m} ΔV=6.0×0.601.20=3.0 V\Delta V = 6.0 \times \frac{0.60}{1.20} = 3.0\,\mathrm{V}

The answer is physically sensible: the two points are half the wire length apart, so they are separated by half the supply p.d.

A sliding contact on the wire turns this into a continuously variable potential divider. Moving the contact changes xx, so it changes the fraction of the supply p.d. selected from the wire. That is the same ratio idea you will use next with fixed resistors.

The Potential Divider Ratio

A potential divider is the same sharing idea, but with two resistors instead of a long uniform wire. The resistors are in series across a supply, and the output p.d. is taken across one chosen resistor.

For two series resistors, the same current flows through both. If the supply p.d. is VsV_{\mathrm{s}}, then:

I=VsR1+R2I = \frac{V_s}{R_1 + R_2}

The p.d. across R2R_2 is:

V2=IR2V_2 = IR_2

Substitute the current:

Potential divider

Vout=VsR2R1+R2V_{\mathrm{out}} = V_{\mathrm{s}} \frac{R_2}{R_1 + R_2}

VsV_{\mathrm{s}}: supply p.d. (V); R2R_2: the resistor the output is taken across; R1R_1: the other series resistor (Ω). Valid when the output draws negligible current (high-resistance voltmeter or input).

The key decision is the numerator. The numerator is the resistance across which the output p.d. is measured. If the output is across R1R_1, use R1R_1 in the numerator. If the output is across R2R_2, use R2R_2.

This simple formula assumes the output is measured by a high-resistance voltmeter or control input, so almost no extra current is drawn from the divider. If you connect a low-resistance load across the output, that load is effectively in parallel with the output resistor, so the output p.d. is no longer given by the unloaded divider ratio.

Worked route:

A 12.0 V12.0\,\mathrm{V} supply is connected across a 2.0 kΩ2.0\,\mathrm{k\Omega} resistor in series with a 4.0 kΩ4.0\,\mathrm{k\Omega} resistor. The output is measured across the 4.0 kΩ4.0\,\mathrm{k\Omega} resistor.

Decision point: the output is across the 4.0 kΩ4.0\,\mathrm{k\Omega} resistor, so 4.0 kΩ4.0\,\mathrm{k\Omega} goes in the numerator.

Vout=12.0×4.02.0+4.0V_{\text{out}} = 12.0 \times \frac{4.0}{2.0 + 4.0} Vout=8.0 VV_{\text{out}} = 8.0\,\mathrm{V}

The larger resistor gets the larger share of the supply p.d. because the same current passes through both resistors and V=IRV = IR.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account