# Determining Internal Resistance ## Experimental Aims - The goal of this practical is to investigate how e.m.f relates to internal resistance by measuring how current and voltage vary when a variable resistor is adjusted - This experiment allows you to determine both the e.m.f and internal resistance of a power supply from graphical analysis ## Variables - **Independent variable:** Resistance of the variable resistor, $R$ (Ω) - **Dependent variables:** Terminal potential difference, $V$ (V) and current, $I$ (A) - **Control variables:** - E.m.f of the power supply (should remain constant) - Internal resistance of the power supply (should remain constant throughout) ## Equipment Required | Apparatus | Purpose | |-----------|---------| | Power supply (e.g. 1.5 V cell) | Provides e.m.f to the circuit | | Fixed resistor | Acts as the internal resistance in series with the cell | | 100 Ω variable resistor | Allows adjustment of current and voltage readings throughout the circuit | | Voltmeter (0–2 V range) | Measures terminal p.d across the load | | Ammeter (0–200 mA range) | Measures circuit current | | Connecting wires | At least 6 leads for electrical connections | | Switch | Opens circuit between readings to prevent battery drain | - **Resolution of instruments:** - Voltmeter: 1 mV - Ammeter: 0.1 mA

Use fairly new cells to ensure the internal resistance doesn't change during measurements

## Experimental Method 1. Connect the power supply and fixed resistor (representing internal resistance $r$) in series—treat these together as a single cell unit 2. With the circuit switch open, record the voltmeter reading $V$—this gives the e.m.f when no current flows 3. Set the variable resistor to its maximum resistance value, then close the switch and quickly record both $V$ and $I$ 4. Open the switch immediately after recording each reading to prevent the cell from heating up and changing its properties 5. Adjust the variable resistor to different values across its full range and record corresponding $V$ and $I$ values 6. Collect at least 8-10 data pairs to ensure reliable graphical analysis 7. Take 3 repeat readings for each resistance setting and calculate mean values to reduce random uncertainty ### Example Data Table | **Resistance of Variable Resistor** $R$/Ω | **Voltmeter Reading** | | | **Ammeter Reading** | | | **Mean** | | |---|---|---|---|---|---|---|---|---| | | 1st | 2nd | 3rd | 1st | 2nd | 3rd | | | | | $V$/V | $V$/V | $V$/V | $I$/mA | $I$/mA | $I$/mA | $V$/V | $I$/mA | | 0 | | | | | | | | | | 10 | | | | | | | | | | 20 | | | | | | | | | | 30 | | | | | | | | | | 40 | | | | | | | | | | 50 | | | | | | | | | | 60 | | | | | | | | | | 70 | | | | | | | | | | 80 | | | | | | | | | | 90 | | | | | | | | | | 100 | | | | | | | | | ## Data Analysis - The fundamental equation linking e.m.f and internal resistance is: $$\varepsilon = I(R + r)$$ - Where: - $\varepsilon$ = electromotive force (V) - $I$ = current (A) - $R$ = load resistance (Ω) - $r$ = internal resistance (Ω) - This can be expanded and rearranged: $$\varepsilon = IR + Ir = V + Ir$$ - Rearranging for terminal p.d $V$: $$V = -rI + \varepsilon$$ - This equation has the form of a straight line: $y = mx + c$

Comparing $V = -rI + \varepsilon$ with $y = mx + c$: gradient = $-r$, y-intercept = $\varepsilon$

- **Graphical analysis:** - Plot terminal p.d ($V$) on the y-axis against current ($I$) on the x-axis - Draw a best-fit straight line through the data points - The **y-intercept** gives the e.m.f of the cell - The **gradient** (which is negative) gives the negative internal resistance: gradient = $-r$ - Therefore: $r = -\text{gradient}$ - To calculate the gradient: - Draw the largest possible right-angled triangle using the line of best fit - Use: gradient = $\frac{\Delta y}{\Delta x} = \frac{\text{change in voltage}}{\text{change in current}}$ ### Example Graph: Terminal P.D vs Current The graph below illustrates the typical linear relationship you should observe. Notice the negative gradient and the y-intercept representing the e.m.f:

## Evaluation & Error Reduction ### Systematic Errors - **Battery heating during measurements:** - Only close the switch long enough to record each voltage-current pair - This prevents the internal resistance from increasing due to temperature changes - Open the switch between all readings to allow cooling ### Random Errors - **Old or depleted cells:** - Use fresh, new cells as the e.m.f and internal resistance of run-down batteries fluctuate unpredictably during use - **Fluctuating readings:** - Wait for both the voltmeter and ammeter readings to stabilize (stop changing) before recording values - **Measurement precision:** - Take at least 3 repeat readings at each resistance value - Calculate the mean of repeats to reduce the impact of random variation

Comparing the measured gradient with the manufacturer's stated resistance value validates your results

## Safety Considerations - This is a low-risk practical, but electrical components can overheat with prolonged use - If you smell burning or notice excessive heat, switch off the power supply immediately - Keep liquids away from all electrical equipment to prevent short circuits and damage - Ensure connections are secure to avoid sparks or loose wiring ## Worked Example: Determining Internal Resistance from Data A student investigates a power supply using a variable load resistor. The ammeter measures current through the circuit and the voltmeter measures terminal p.d for different resistance values. The collected data is: | $V$/V | $I$/mA | |-------|--------| | 4.60 | 0.0 | | 4.30 | 13.1 | | 1.14 | 20.0 | | 0.96 | 28.2 | | 0.80 | 35.2 | | 0.65 | 41.5 | | 0.49 | 48.8 | | 0.38 | 53.4 | | 0.23 | 60.3 | | 0.10 | 66.0 | Determine the e.m.f and internal resistance from a graph. ### Solution **Step 1: Plot the graph** - Plot $V$ (y-axis) against $I$ (x-axis, converted to mA for easier reading) - Draw a line of best fit through the points The experimental data is plotted below:

**Step 2: Calculate the gradient** - Draw the largest triangle possible along the best-fit line to maximize accuracy - From the graph: - $\Delta V = 1.6 - 0.1 = 1.5$ V - $\Delta I = 0 - 66 = -66$ mA = $-66 \times 10^{-3}$ A - Gradient = $\frac{\Delta V}{\Delta I} = \frac{1.5}{-66 \times 10^{-3}} = -22.7$ Ω (to 3 s.f.)

A negative gradient indicates that terminal p.d decreases as current increases due to energy lost in internal resistance

**Step 3: Extract values from the graph equation** - From $V = -rI + \varepsilon$: - Gradient = $-r$ - Y-intercept = $\varepsilon$ - Therefore: - **Internal resistance:** $r = 22.7$ Ω - **E.m.f:** $\varepsilon = 1.60$ V - The y-intercept is read directly from where the best-fit line crosses the y-axis at $I = 0$