A loaded potential divider: why the calculated voltage was wrong
A student must provide 4.0 V across a 200 Ω sensor from a 9.0 V PP3 battery of internal resistance 1.5 Ω. Using the E12 series, the student first chooses R1 = 1.0 kΩ in series with the battery and R2 = 820 Ω in parallel with the sensor. The unloaded divider would give 4.05 V, but when the sensor is connected the measured voltage is only 1.25 V. The sensor resistance changes by ±10% as it warms. The student has 0.25 W resistors and needs the sensor voltage to stay within ±5% of 4.0 V. The available E12 values include 82 Ω, 100 Ω, 120 Ω, 150 Ω, 180 Ω, 220 Ω and 270 Ω. Decide whether the original design can be salvaged, select a better pair, and assess the effect of the battery’s internal resistance.
1. Explain why the measured voltage is 1.25 V rather than the predicted 4.05 V.
Read the worked answer
The prediction 4.05 V is the unloaded divider voltage:
V_out = 9.0 × 820/(1000 + 820) = 4.05 V.
But the sensor is not a high-resistance voltmeter. It is a 200 Ω load in parallel with R2, so the lower arm becomes
R_p = (820 × 200)/(820 + 200) = 160.8 Ω.
The loaded output is therefore
V_out = 9.0 × 160.8/(1000 + 160.8) = 1.25 V.
The 820 Ω resistor is too large compared with the 200 Ω sensor: most of the current that would have flowed through R2 is diverted through the sensor, pulling the output voltage down. The original design is not a stable 4 V source for this load.
2. Select a suitable E12 pair and check that the resistor power ratings and the ±10% sensor variation are acceptable.
Read the worked answer
For a loaded divider, first combine R2 with the 200 Ω sensor:
R_p = 200R2/(200 + R2).
The required ratio is V_out/9.0 = 4.0/9.0, so
R_p/(R1 + R_p) = 4/9,
which gives R1 = 1.25R_p.
Try R2 = 180 Ω:
R_p = (180 × 200)/(380) = 94.7 Ω,
R1 = 1.25 × 94.7 = 118 Ω.
The nearest E12 value is 120 Ω. Then
V_out = 9.0 × 94.7/(120 + 94.7) = 3.97 V.
The total current is about 9.0/214.7 = 41.9 mA. The power in R1 is I²R1 = (0.0419)²(120) = 0.211 W, below 0.25 W. The power in R2 is V_out²/R2 = 3.97²/180 = 0.088 W, also safe.
Check the sensor extremes. If the sensor becomes 180 Ω, R_p = 90.0 Ω and V_out = 9.0 × 90.0/(210) = 3.86 V, a 3.6% drop. If it becomes 220 Ω, R_p = 99.0 Ω and V_out = 9.0 × 99.0/(219) = 4.07 V, a 1.7% rise. Both are within ±5%. A lower-resistance pair such as 82 Ω and 100 Ω would give a similar voltage but would exceed the 0.25 W rating in R1, so 120 Ω and 180 Ω is the better choice.
3. Assess the effect of the battery’s internal resistance and whether the original design can be salvaged without active components.
Read the worked answer
With the 120 Ω and 180 Ω pair, the total resistance is about 214.7 Ω plus the internal 1.5 Ω. The current is 9.0/216.2 = 41.6 mA, so the internal drop is only 0.062 V. The terminal voltage is 8.94 V and the output becomes about 3.94 V, a further 0.7% reduction. This is small compared with the ±5% tolerance and does not upset the design. Internal resistance would matter if the divider resistors were comparable with 1.5 Ω, which they are not.
The original 1 kΩ/820 Ω design cannot be salvaged as a passive divider for a 200 Ω load: the output is far too low because the lower arm is dominated by the sensor. Adding resistors in parallel with the sensor would change the load and is not a proper fix; an active buffer could be used, but that is outside the passive divider design. A passive redesign with much smaller resistors is required, and the chosen 120 Ω/180 Ω pair meets the voltage, tolerance and power constraints.