Current Electricity – Study Notes
(Physics – Bachelor of Physiotherapy, BPT)
1. Simple Explanation
Current electricity is the flow of electric charge (usually electrons) through a conductor.
- Electric current (I) is the amount of charge that crosses a cross‑section per unit time.
- Voltage (V) (or electromotive force, EMF) is the driving force that pushes charges.
- Resistance (R) is the opposition to that flow.
The fundamental relationship between these three quantities is given by Ohm’s Law:
[
V = I , R
]
2. Key Concepts & Points
| Concept | Symbol | Units | Typical Values | Notes |
|---|---|---|---|---|
| Current | I | Amperes (A) | 0.001 A – 10 A (in medical devices) | 1 A = 1 C/s |
| Voltage | V | Volts (V) | 0 – 100 V (low‑voltage medical devices) | 1 V = 1 J/C |
| Resistance | R | Ohms (Ω) | 1 Ω – 10 kΩ | Depends on material, length, cross‑section |
| Conductance | G | Siemens (S) | G = 1/R | Inverse of resistance |
| Power | P | Watts (W) | P = VI = I²R = V²/R | Energy per unit time |
2.1 Ohm’s Law
- V = I R ⇔ I = V/R ⇔ R = V/I
- Valid when the material’s resistance is constant (ohmic).
2.2 Resistivity (ρ)
- Material property: ( R = \rho \frac{L}{A} )
- (L) = length of conductor
- (A) = cross‑sectional area
- Lower resistivity → better conductor (e.g., copper, silver).
2.3 Series & Parallel Circuits
- Series: ( R_{\text{total}} = R_1 + R_2 + \dots )
- Same current flows through each element.
- Parallel: ( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots )
- Same voltage across each element.
2.4 Kirchhoff’s Laws
- KCL (Kirchhoff’s Current Law): Sum of currents entering a node = sum leaving.
- KVL (Kirchhoff’s Voltage Law): Sum of voltages around any closed loop = 0.
2.5 Electromotive Force (EMF) & Internal Resistance
- Real batteries have an EMF (E) and internal resistance (r).
- Terminal voltage ( V = E - I r ).
2.6 Power Dissipation
- In a resistor: ( P = I^2 R = V^2 / R ).
- Useful for understanding heating in wires and safety limits.
2.7 Relevance to Physiotherapy
- Electrical Stimulation Devices (TENS, EMS) rely on controlled current/voltage.
- Safety requires knowledge of impedance of human tissue, safe current limits (<2 mA for surface stimulation).
- Low‑voltage, high‑impedance circuits are common in therapeutic units.
3. Worked Example
Problem
A TENS unit delivers a sinusoidal current of 1 mA at a voltage of 5 V through a skin patch.
- Find the resistance of the patch.
- Compute the power dissipated in the patch.
- If the patch area is ( 4 \text{ cm}^2 ) and the patch thickness is ( 0.5 \text{ mm} ), estimate the resistivity of the patch material.
3.1 Solution
Step 1: Resistance (Ohm’s Law)
[
R = \frac{V}{I} = \frac{5\ \text{V}}{1\ \text{mA}} = \frac{5}{0.001} = 5000\ \Omega
]
Step 2: Power Dissipation
[
P = I^2 R = (0.001\ \text{A})^2 \times 5000\ \Omega = 5 \times 10^{-6}\ \text{W} = 5\ \mu\text{W}
]
(Alternatively, (P = V I = 5 \times 0.001 = 5\ \mu\text{W}).)
Step 3: Resistivity
First find geometry:
- Thickness ( t = 0.5\ \text{mm} = 0.0005\ \text{m} )
- Area ( A = 4\ \text{cm}^2 = 4 \times 10^{-4}\ \text{m}^2 )
- Length ( L = t ) (current flows through thickness).
Using ( R = \rho \frac{L}{A} ):
[
\rho = R \frac{A}{L} = 5000\ \Omega \times \frac{4 \times 10^{-4}\ \text{m}^2}{0.0005\ \text{m}} = 5000 \times 0.8 = 4000\ \Omega\cdot\text{m}
]
Interpretation
- The patch behaves like a resistor of 5 kΩ, dissipating only 5 µW, well below safety limits.
- The resistivity