**Question 1**
**APPARATUS:** Standard resistor, Resistance box, Connecting wires,
*(Circuit diagram showing: Battery E, Key K, Ammeter A, Standard resistor X, and Resistance box R connected in a circuit)*
Ammeter (1 – 5)A, Key, Battery E (3V), Potentiometer, Voltmeter.
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**PROCEDURE:**
**(a)** You are provided with a standard resistor X, a resistance box R, connecting wires, an ammeter A, a key K and a battery E.
i. Connect the circuit as shown in the diagram above.
ii. Set the resistance box R = 2Ω and close the key K.
iii. Read and record the value of current I on the ammeter.
iv. Evaluate R⁻¹.
v. Repeat the above procedure for **four** other values of R = 4Ω, 6Ω, 8Ω and 10Ω. In each case, read and record the corresponding values of I. Also evaluate R⁻¹.
# Solution to Question 1
## Understanding the Circuit
The circuit has a battery E (3V), standard resistor X, resistance box R, and ammeter A in series.
Using **Ohm's Law**: E = I(X + R)
Rearranging: **1/I = X/E · (1/R)⁻¹... **
Actually, let's derive properly:
$$E = I(X + R)$$
$$\frac{1}{I} = \frac{X + R}{E} = \frac{X}{E} + \frac{R}{E}$$
So: **I = E/(X + R)**
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## Part (a) — Results Table
Assuming E = 3V and X = 1Ω (standard resistor):
| R (Ω) | R⁻¹ (Ω⁻¹) | I = 3/(1+R) (A) |
|--------|------------|-----------------|
| 2 | 0.500 | 1.000 |
| 4 | 0.250 | 0.600 |
| 6 | 0.167 | 0.429 |
| 8 | 0.125 | 0.333 |
| 10 | 0.100 | 0.273 |
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## Graph (b) — Expected
Plot **I** (y-axis) vs **R⁻¹** (x-axis)
From: $I = \frac{E}{X} \cdot R^{-1} \cdot \frac{1}{1 + \frac{X}{R}}$
More usefully, rearranging:
$$\frac{1}{I} = \frac{X}{E} + \frac{1}{E} \cdot R$$
A graph of **I vs R⁻¹** gives a **straight line through the origin** with:
- **Slope = E/X** (since as R→∞, I→0)
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## Key Relationships
| Quantity | Formula |
|----------|---------|
| Current | I = E/(X + R) |
| R⁻¹ | 1/R |
| Gradient of I vs R⁻¹ | = E (if X is negligible) |
| X from graph | X = E/gradient − intercept |
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## Finding X (Standard Resistor)
From the linear equation:
$$I = \frac{E}{X} R^{-1} \cdot \frac{R}{1}$$
Better form — plot **1/I vs R**:
$$\frac{1}{I} = \frac{R}{E} + \frac{X}{E}$$
- **Gradient = 1/E** → confirms battery EMF
- **Y-intercept = X/E** → so **X = intercept × E**
If intercept = 0.333 and E = 3V → **X = 1Ω** ✓