Potential dividers
Section: D.C. Circuits | Syllabus: Cambridge AS Level Physics 9702
The Potential Divider Principle A potential divider is a circuit that uses two or more resistors in series to divide an input voltage into smaller output voltages. The output voltage depends on the ratio of the resistances.
Potential Divider A circuit arrangement of resistors in series that divides a supply voltage in proportion to the resistance values. FIG 10.11: Basic Potential Divider Circuit Show a circuit with a voltage source V_in connected to two resistors R₁ and R₂ in series.
The output voltage V_out is taken across R₂ (between the junction of R₁ and R₂, and the negative terminal). Label V_in across the whole circuit and V_out across R₂. Show a high-resistance voltmeter connected across R₂ to measure V_out.
How It Works Since R_1 and R_2 are in series: The same current I flows through both resistors The total resistance is R_T = R_1 + R_2 The input voltage divides between the resistors in proportion to their resistances Applying V = IR to each resistor: V_1 = IR_1 and V_2 = IR_2 The ratio of voltages equals the ratio of resistances: fraction = fraction = fraction The Potential Divider Equation The output voltage across R_2 can be expressed as a fraction of the input voltage: Potential Divider Equation V_out = fraction × V_in where V_out is measured across R_2 Derivation The current through the circuit is: I = V_inR_1 + R_2 The voltage across R_2 is: V_out = IR_2 = V_inR_1 + R_2 × R_2 = fraction × V_in Key Insight The output voltage is always a fraction of the input voltage.
The larger the proportion of total resistance that R_2 represents, the larger the fraction of input voltage that appears across it. Worked Example: Basic Potential Divider Problem: A 9 V battery is connected to a potential divider with R_1 = 700\, and R_2 = 300\,.
Calculate the output voltage across R_2. Solution: V_out = fraction × V_in V_out = fraction × 9 V_out = fraction × 9 = 0.3 × 9 = 2.7 V Assumptions in Ideal Potential Dividers The voltmeter measuring V_out has infinite resistance (draws no current) No current is drawn from the output terminals by any connected circuit If current is drawn from the output, the output voltage will be less than calculated Sensors in Potential Dividers Potential dividers become particularly useful when one resistor is replaced by a sensor whose resistance varies with environmental conditions.
This creates an output voltage that depends on temperature or light intensity. Thermistors (NTC Type) An NTC (Negative Temperature Coefficient) thermistor has a resistance that decreases as temperature increases .
NTC Thermistor A temperature-sensitive resistor whose resistance decreases non-linearly with increasing temperature. FIG 10.12: Thermistor in a Potential Divider Show two circuits side by side: Circuit A: Thermistor at the top (in R₁ position), fixed resistor at bottom.
Label "V_out increases as temperature rises" because the thermistor resistance decreases, so more voltage appears across the fixed resistor. Circuit B: Fixed resistor at top, thermistor at bottom (in R₂ position).
Label "V_out decreases as temperature rises" because the thermistor resistance (R₂) decreases, reducing V_out. Worked Example: Thermistor Potential Divider Problem: A thermistor has resistance 2 kΩ at 20°C and 500 Ω at 50°C.
It is connected in series with a 1 kΩ fixed resistor across a 9 V supply. The output is taken across the fixed resistor. Calculate the output voltage at (a) 20°C and (b) 50°C. Solution: (a) At 20°C: Thermistor resistance = 2 kΩ V_out = fraction × 9 = fraction × 9 = 3 V (b) At 50°C: Thermistor resistance = 500 Ω V_out = fraction × 9 = fraction × 9 = 6 V The output voltage increases from 3 V to 6 V as temperature rises, because the thermistor resistance decreases.
Light-Dependent Resistors (LDRs) An LDR has a resistance that decreases as light intensity increases . Light-Dependent Resistor (LDR) A light-sensitive resistor whose resistance decreases as the intensity of light falling on it increases.
Resistance can range from ~100 Ω in bright light to ~1 MΩ in darkness. FIG 10.13: LDR in a Potential Divider Show two configurations: Configuration 1: LDR at top, fixed resistor at bottom - V_out increases in bright light (for circuits that activate in light).
Configuration 2: Fixed resistor at top, LDR at bottom - V_out decreases in bright light (for circuits that activate in darkness). Include a note showing that this can control automatic lighting systems.
Worked Example: LDR Potential Divider Problem: An LDR has resistance 230 Ω in daylight and 1 MΩ in darkness. It is placed in series with a 10 kΩ resistor across a 9 V supply. The output is taken across the LDR.
Calculate V_out in (a) daylight and (b) darkness. Solution: (a) In daylight: LDR resistance = 230 Ω V_out = fraction × 9 = fraction × 9 = 0.20 V (b) In darkness: LDR resistance = 1 MΩ = 1,000,000 Ω V_out = fraction × 9 = fraction × 9 = 8.9 V The output across the LDR is low in daylight and high in darkness.
Control Circuits and Automatic Sy…
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