Potential dividers

1. Overview

The potential divider principle is a direct application of Kirchhoff’s Second Law and Ohm’s Law to series circuits. It describes how the total electromotive force (e.m.f.) of a power supply is distributed across components in proportion to their resistances. This principle allows for the creation of variable output voltages, the design of environmental sensors (using thermistors and LDRs), and the high-precision measurement of e.m.f. through the use of potentiometers and null methods.

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Key Definitions

• Potential Divider: A simple circuit that uses two or more resistors in series to split the input voltage into a specific ratio, providing a fraction of the source voltage as an output.
• Potentiometer: A precision instrument consisting of a uniform resistance wire and a sliding contact (jockey). It acts as a continuously variable potential divider and is used to measure or compare potential differences without drawing current from the source being measured.
• Null Method: An experimental technique where a circuit is adjusted until a galvanometer shows zero deflection. This indicates a state of equilibrium where the potential difference across two points is exactly equal, meaning no current flows between them.
• Galvanometer: A highly sensitive instrument designed to detect and measure minute electric currents. In potential divider circuits, it is primarily used to identify the null point (zero current).
• Light-Dependent Resistor (LDR): A semiconductor device whose resistance decreases non-linearly as the light intensity incident upon it increases.
• Negative Temperature Coefficient (NTC) Thermistor: A sensor whose resistance decreases as its temperature increases, commonly used in temperature-sensing potential dividers.
• Potential Gradient: The change in potential per unit length along a uniform resistance wire, typically expressed in $\text{V m}^{-1}$.

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Content

3.1 The Potential Divider Principle

In a series circuit, the current $I$ is constant through all components. According to Ohm’s Law ($V = IR$), the potential difference across any resistor is directly proportional to its resistance.

Mathematical Derivation: Consider a circuit with an input voltage $V_{in}$ and two resistors $R_1$ and $R_2$ in series. We wish to find the output voltage $V_{out}$ across $R_2$.

1. Total Resistance ($R_T$): $$R_T = R_1 + R_2$$
2. Circuit Current ($I$): $$I = \frac{V_{in}}{R_T} = \frac{V_{in}}{R_1 + R_2}$$
3. Voltage across $R_2$ ($V_{out}$): $$V_{out} = I \times R_2$$
4. Substitution: Substituting the expression for $I$ into the $V_{out}$ equation: $V_{out} = V_{in} \left( \frac{R_2}{R_1 + R_2} \right)$

The Ratio Rule: A more intuitive way to view potential dividers is through the ratio of voltages: $\frac{V_1}{V_2} = \frac{R_1}{R_2}$ This shows that the larger the resistance, the larger the share of the total potential difference it receives.

3.2 Sensor Circuits: Thermistors and LDRs

By replacing one of the fixed resistors in a potential divider with a transducer (a component that changes resistance in response to environmental stimuli), we can create a circuit where $V_{out}$ responds to temperature or light.

• NTC Thermistor (Temperature Sensor):

  • Mechanism: As temperature rises, more charge carriers are liberated in the semiconductor, so resistance $R$ decreases.
  • Circuit Behavior: If the thermistor is $R_2$ (the output component), an increase in temperature leads to a decrease in $V_{out}$.
  • Inversion: To create a circuit where $V_{out}$ increases with temperature (e.g., to trigger a cooling fan), the thermistor must be placed in the $R_1$ position, or $V_{out}$ must be taken across the fixed resistor $R_2$.

• Light-Dependent Resistor (LDR) (Light Sensor):

  • Mechanism: As light intensity increases, photon energy releases more electrons, so resistance $R$ decreases.
  • Circuit Behavior: If the LDR is $R_2$, $V_{out}$ decreases as the environment gets brighter.
  • Application: This is used in "dark-activated" switches. When it gets dark, $R_{LDR}$ increases, $V_{out}$ across the LDR increases, and this voltage can trigger a transistor or relay to turn on a lamp.

3.3 The Potentiometer

A potentiometer is a specialized potential divider. It uses a long, uniform wire (often $1.00\text{ m}$ long) made of a high-resistivity alloy like constantan.

• Uniformity: Because the wire has a uniform cross-sectional area and composition, its resistance is directly proportional to its length ($R \propto L$).
• Potential Gradient ($k$): The potential difference per unit length along the wire is constant. $k = \frac{V_{wire}}{L_{wire}}$ Therefore, the potential difference $V_x$ across any length $L_x$ is: $V_x = k \times L_x = \left( \frac{V_{wire}}{L_{wire}} \right) L_x$

3.4 Comparing Potential Differences and Null Methods

The primary advantage of a potentiometer is its ability to measure e.m.f. without drawing any current from the source being measured. This is achieved using a null method.

The Experimental Setup:

1. Driver Circuit: A driver cell (e.m.f. $E_D$) is connected across the potentiometer wire $AB$. This creates a potential gradient.
2. Test Circuit: The cell to be measured ($E_x$) is connected in parallel to a section of the wire. A galvanometer is placed in series with $E_x$.
3. Polarity: The positive terminal of $E_x$ must be connected to the same end of the wire as the positive terminal of the driver cell.
4. Balance Point: The jockey is moved along the wire until the galvanometer reads zero. At this "null point," the potential difference across the length of wire $L_x$ is exactly equal to $E_x$.

Why is this more accurate than a voltmeter?

• A standard voltmeter has a high, but finite, resistance. It must draw a small current to operate. This current causes "lost volts" ($Ir$) across the internal resistance of the cell being measured, so the voltmeter measures terminal P.D. ($V$), not e.m.f. ($E$).
• At the null point of a potentiometer, $I = 0$ in the test circuit. Since $V = E - Ir$ and $I = 0$, then $V = E$. The potentiometer measures the true e.m.f.

Comparing Two Cells: If two cells with e.m.f.s $E_1$ and $E_2$ have balance lengths $L_1$ and $L_2$ on the same potentiometer: $\frac{E_1}{E_2} = \frac{L_1}{L_2}$

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4. Worked Examples

Worked Example 1 — Sensor Circuit Design

Question: A $12.0\text{ V}$ supply with negligible internal resistance is connected to a fixed resistor of $4.0\text{ k}\Omega$ and an LDR in series. The output voltage $V_{out}$ is measured across the $4.0\text{ k}\Omega$ resistor. In bright sunlight, the LDR resistance is $800\text{ }\Omega$. In the shade, the LDR resistance is $10.0\text{ k}\Omega$. Calculate the change in $V_{out}$ when the LDR is moved from sunlight to shade.

Step 1: Calculate $V_{out}$ in sunlight The LDR is $R_1$ and the fixed resistor is $R_2 = 4000\text{ }\Omega$. $$V_{out(sun)} = V_{in} \left( \frac{R_2}{R_1 + R_2} \right)$$ $$V_{out(sun)} = 12.0 \left( \frac{4000}{800 + 4000} \right)$$ $$V_{out(sun)} = 12.0 \left( \frac{4000}{4800} \right) = 10.0\text{ V}$$

Step 2: Calculate $V_{out}$ in shade $$V_{out(shade)} = 12.0 \left( \frac{4000}{10000 + 4000} \right)$$ $$V_{out(shade)} = 12.0 \left( \frac{4000}{14000} \right) \approx 3.43\text{ V}$$

Step 3: Calculate the change $$\Delta V_{out} = 10.0\text{ V} - 3.43\text{ V} = 6.57\text{ V}$$ Answer: The output voltage decreases by $6.57\text{ V}$.

Worked Example 2 — Potentiometer with Internal Resistance

Question: A potentiometer wire $AB$ has a length of $100\text{ cm}$ and a resistance of $8.0\text{ }\Omega$. It is connected to a driver cell of e.m.f. $4.0\text{ V}$ and internal resistance $2.0\text{ }\Omega$. A test cell of e.m.f. $E_x$ gives a balance point at $AC = 70.0\text{ cm}$. Calculate $E_x$.

Step 1: Calculate the total resistance of the driver circuit $$R_{total} = R_{wire} + r = 8.0 + 2.0 = 10.0\text{ }\Omega$$

Step 2: Calculate the potential difference across the wire ($V_{AB}$) Using the potential divider principle for the driver circuit: $$V_{AB} = E_{driver} \left( \frac{R_{wire}}{R_{total}} \right)$$ $$V_{AB} = 4.0 \left( \frac{8.0}{10.0} \right) = 3.2\text{ V}$$

Step 3: Calculate the potential gradient ($k$) $$k = \frac{V_{AB}}{L_{AB}} = \frac{3.2\text{ V}}{100\text{ cm}} = 0.032\text{ V cm}^{-1}$$

Step 4: Calculate $E_x$ using the balance length $$E_x = k \times L_{balance}$$ $$E_x = 0.032\text{ V cm}^{-1} \times 70.0\text{ cm} = 2.24\text{ V}$$ Answer: $E_x = 2.24\text{ V}$

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Key Equations

Equation	Description	Data Sheet?
$V_{out} = V_{in} \left( \frac{R_2}{R_1 + R_2} \right)$	Potential divider formula for $V$ across $R_2$.	No
$\frac{V_1}{V_2} = \frac{R_1}{R_2}$	Ratio of voltages in a series circuit.	No
$k = \frac{V}{L}$	Potential gradient of a uniform wire.	No
$E_x = \frac{L_x}{L_{total}} V_{wire}$	Finding unknown e.m.f. with a potentiometer.	No
$\frac{E_1}{E_2} = \frac{L_1}{L_2}$	Comparing two e.m.f.s using balance lengths.	No

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Common Mistakes to Avoid

• ❌ Wrong: Using the e.m.f. of the driver cell as $V_{in}$ for the potentiometer wire when the driver cell has internal resistance. ✓ Right: Always calculate the actual terminal P.D. across the wire first using $V = E - Ir$ or the potential divider formula.
• ❌ Wrong: Assuming $V_{out}$ is always across the variable sensor. ✓ Right: Read the diagram carefully. If $V_{out}$ is across the fixed resistor, its value will increase when the sensor's resistance decreases.
• ❌ Wrong: Connecting the test cell with the wrong polarity. ✓ Right: The positive terminal of the test cell must face the positive terminal of the driver cell. If they oppose each other, the galvanometer will never reach zero.
• ❌ Wrong: Trying to measure a test e.m.f. that is larger than the P.D. across the potentiometer wire. ✓ Right: If $E_x > V_{wire}$, the balance point will be "off the scale" (beyond the length of the wire).
• ❌ Wrong: Forgetting to convert units (e.g., mixing cm and m). ✓ Right: Ensure all lengths are in the same units before calculating ratios.

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Exam Tips

1. The "No Current" Argument: In any question asking why a potentiometer is superior to a voltmeter, the phrase "draws no current at the balance point" is almost always a required marking point.
2. Sensitivity: To increase the precision of a potentiometer (make it more sensitive), you can decrease the potential gradient. This is done by increasing the length of the wire or adding a resistor in series with the driver cell to reduce $V_{wire}$.
3. Logical Flow for Sensors: When explaining how a sensor circuit works, follow this 3-step logic:
   • 1. Environmental change $\rightarrow$ Resistance change (e.g., "Temp $\uparrow$, $R_{thermistor} \downarrow$").
   • 2. Resistance change $\rightarrow$ Total resistance and current change (or ratio change).
   • 3. Ratio change $\rightarrow$ $V_{out}$ change (referencing the potential divider equation).
4. Loading Effect: Be aware that if a "load" (like a motor or a low-resistance voltmeter) is connected in parallel with $R_2$, the effective resistance of that branch decreases, which in turn decreases $V_{out}$.
5. Potentiometer Uniformity: If a question mentions the wire is "non-uniform," the relationship $V \propto L$ no longer holds, and the potential gradient is not constant.