Notes on Resistance, Ohm's Law, and Circuits
Resistance: Definition and Basic Relationships
Resistance is defined as an obstacle to the flow of electric current. It is the opposition offered by any object to the passage of an electric current through it.
Fundamental idea: resistance quantifies how hard it is for charges to move through a material.
Length as a Determinant of Resistance
Resistance of a conductor is directly proportional to its length: longer wire → greater resistance; shorter wire → smaller resistance.
If L represents the length of a uniform wire, then the resistance varies with length asR \propto L.
For a uniform wire, the relation between resistances of two lengths can be expressed {R1}/{R2}={L1}/{L2}
where R1 and R2 are resistances and L1 and L2 are the corresponding lengths.Explanation: Doubling the length doubles the number of obstacles (scattering sites, collisions) encountered by moving charges, increasing resistance.
Resistivity and the Core Formula for Resistance
Resistivity (ρ) is the resistance per unit length of a specific substance; it is a material constant (for a given temperature).
For a uniform conductor of length L and cross-sectional area A, resistance is given byR=\frac{\rho L}{A}.
Implication: both longer length (increases R) and smaller cross-sectional area (decreases A, increases R) raise resistance; larger cross-sectional area lowers resistance.
Practical Illustration: Volume Control as a Variable Resistor
A volume control knob acts as a variable resistor: as the knob turns, the effective length (or path) of resistance changes.
Effect: changing the resistance alters the current through the circuit, thereby changing the loudness of the speaker.
Diameter (Cross-Sectional Area) and Resistance
Resistance is inversely proportional to the cross-sectional area A: thicker wire (larger A) → smaller resistance; thinner wire (smaller A) → larger resistance.
For a wire of cross-sectional area A, the relation with two wires is {R1}/{R2}={A2}/{A1}.
If the cross-sectional area is doubled, twice as many electrons are available to flow, so the current is doubled (for the same material and length).
The wire gauge number indicates its size: smaller gauge number → thicker wire → lower resistance.
Area, Radius, and a Worked Cross-Section Example
If a wire’s diameter is d, then cross-sectional area isA=\pi r^2=\pi(\tfrac{d}{2})^2.
Example problem (radius and area): given d, compute r and A to determine R viaR=\frac{\rho L}{A}.
Note: Stranded wires effectively increase the cross-sectional area, reducing resistance similarly to increasing diameter.
Worked Problem: Resistance vs. Diameter
Given an iron wire with diameter d1=0.8 mm, resistance R1=0.4 Ω.
Find R2 for a wire with d2=0.4 mm (same material and length).
Since R ∝ 1/d^2 (because A ∝ d^2), we use {R1}/{R2}={d2^2}/{d1^2}.
Calculation:
d1^2 = (0.8 mm)^2 = 0.64 mm^2
d2^2 = (0.4 mm)^2 = 0.16 mm^2
R2 = R1 × (d1^2 / d2^2) = 0.4 × (0.64 / 0.16) = 0.4 × 4 = 1.6 Ω.
Temperature Effects on Resistance
For metallic conductors, resistance increases as temperature increases (example: copper).
For semiconductors/insulators, resistance decreases as temperature increases.
A rough sense: as wires get warmer, their resistance typically increases; colder wires have lower resistance.
Temperature trends can be summarized with a qualitative note: Cold wire vs. Warm wire graphs show higher resistance when warm.
Material Resistivity and the Role of ρ
Resistivity ρ is a material constant that determines how strongly a material opposes current, independent of geometry.
For a conductor of length L and cross-sectional area A, the resistance isR=\frac{\rho L}{A}.
Key takeaway: materials with lower ρ conduct better (lower resistance) for the same geometry and temperature.
Selected Resistivity Values (Table Overview)
Metals:
Silver: {\rho}=1.60\times 10^{-8}\ \Omega\cdot m
Copper: {\rho}=1.62\times 10^{-8}\ \Omega\cdot m
Aluminium: {\rho}=2.63\times 10^{-8}\ \Omega\cdot m
Tungsten: {\rho}=5.20\times 10^{-8}\ \Omega\cdot m
Nickel: {\rho}=6.84\times 10^{-8}\ \Omega\cdot m
Iron: {\rho}=1.00\times 10^{-7}\ \Omega\cdot m
Chromium: {\rho}=1.29\times 10^{-7}\ \Omega\cdot m
Mercury: {\rho}=9.40\times 10^{-7}\ \Omega\cdot m
Alloys and others:
Manganese: {\rho}=1.84\times 10^{-6}\ \Omega\cdot m
Constantan (Cu-Ni): {\rho}=4.9\times 10^{-5}\ \Omega\cdot m
Manganin (Cu-Mn-Ni): {\rho}=4.4\times 10^{-5}\ \Omega\cdot m
Nichrome (Ni-Cr-Mn-Fe): {\rho}=1.00\times 10^{-4}\ \Omega\cdot m
Insulators (high resistivity):
Glass: around 10^12–10^14 Ω·m (typical insulating range)
Hard rubber: around 10^13–10^16 Ω·m
Ebonite: around 10^15–10^17 Ω·m
Diamond: around 10^12–10^13 Ω·m
Paper (dry): around 10^12 Ω·m
Note: The transcript lists a column of resistivity values for materials, including several insulators, with high resistivity compared to metals.
Problem: Given L, diameter, and ρ, Find Resistance
Given: L = 15 m, diameter d = 0.085 cm, ρ = 1.6×10^-8 Ω·m, T = 20°C.
Conversion: d = 0.085 cm → m: d = 8.5×10^-4 m; radius r = d/2 = 4.25×10^-4 m.
Cross-sectional area: A = \pi r^2 = \pi (4.25\times 10^{-4})^2 = 5.67\times 10^{-7}\ \mathrm{m^2}.
Resistance: R = \frac{\rho L}{A} = \frac{(1.6\times 10^{-8}) (15)}{5.67\times 10^{-7}} \approx 0.42\text{–}0.49\ \Omega.
The transcript reports a final value of R = 0.492\ \Omega.
Steps summary:
Convert dimensions to SI units.
Compute cross-sectional area from diameter.
Apply $R=\rho L/A$.
Ohm’s Law and Basic Electrical Quantities
Ohm’s Law: At steady state, the voltage across a resistor is directly proportional to the current through it at constant temperature:V = IR.
Symbols:
$V$: applied voltage (volts)
$I$: current (amperes)
$R$: resistance (ohms)
Other fundamental quantities:
Electric Voltage (V or E): driving force behind current; unit is the Volt (V).
Electric Current (I): motion/flow of charges; unit is the Ampere (A).
Electric Potential Difference: unit is the Volt.
Electric Resistance (R): limits current; unit is the Ohm (Ω).
Ohm's Law: Conditions and Limitations
Conditions for Ohm's Law:
It can be applied to the entire circuit or to a part of a circuit (use the resistance and voltage drop across that part for the part-circuit application).
It applies to both DC and AC circuits (with appropriate interpretation for AC impedance).
Limitations (non-ohmic behavior): Ohm’s law is not applicable to:
Metals that heat up significantly as current flows (temperature dependence).
Electrolytes with large gas production at electrodes.
Vacuum tubes, arc lamps, and some semiconductor devices.
Gas-filled tubes where ions are generated by current flow.
Devices whose operation depends on current direction (diodes, certain rectifiers, detectors).
Electrical Power and Energy
Electrical Power (P): rate at which electrical energy is expended or converted.
Units: watt (W).
Primary formulas:P = VI = I^2R = \frac{V^2}{R}.
Relationship to energy: Energy (E) is power over time, E = Pt.
Common equivalences: 1 horsepower (HP) = 746 W.
Fundamental Electrical Quantities and Units
Electric Voltage: driving force for current; unit Volt (V).
Electric Current: movement of charges; unit Ampere (A).
Electric Potential Difference: difference in potentials; unit Volt (V).
Electric Resistance: limits current flow; unit Ohm (Ω).
Energy and Power Calculations in Everyday Terms
Energy is the capacity to do work; unit: Joule (J).
Electrical energy consumption is the rate of energy use over time; unit often expressed in kilowatt-hours (kWh).
Example derivation: If a device uses power $P$ for time $t$, energy consumed is W = P\,t. (Note: $W$ often stands for energy in joules, while electrical energy billed uses kWh.)
Electric Circuits: Components and Modes
An electric circuit is a collection of electrical elements interconnected so that current can flow (or is intended to flow).
A circuit typically includes: a source (voltage or current), conductors, and a load.
Circuit States: Closed, Open, and Short Circuits
Closed Circuit: uninterrupted path allowing current to flow; switch closed/on.
Open Circuit: path is interrupted; current does not flow.
Short Circuit: unintended current path bypasses the load; current flows through an unintended path due to damaged insulation or a fault.
Circuit Connections: Series, Parallel, and Combinational
Series Circuit: components connected end-to-end; same current flows through all elements.
Parallel Circuit: components connected across the same two nodes; currents split among branches.
Combinational Circuits: combinations of series and parallel arrangements.
Series-parallel: a simplified form that results in a series circuit upon reduction.
Parallel-series: a simplified form that results in a parallel circuit upon reduction.
Series Circuits: Properties and Calculations
In a series circuit:
The same current flows through all resistors.
There are voltage drops across each resistor.
The equivalent resistance is the sum.
Notation used in the example:
$RT$ = total resistance, $IT$ = circuit current, $V_T$ = total applied voltage.
Relationship: VT = IT R_T.
Series Circuit Sample Problem 1
Problem: Four coils with resistances 3 Ω, 5 Ω, 10 Ω, 12 Ω are connected in series across 120 V.
(a) Equivalent resistance: R_T = 3 + 5 + 10 + 12 = 30\ \Omega.
(b) Current: IT = \frac{VT}{R_T} = \frac{120}{30} = 4\ A.
(c) Voltage drop across each coil:
V1 = IT R_1 = 4\times 3 = 12\ V,
V_2 = 4\times 5 = 20\ V,
V_3 = 4\times 10 = 40\ V,
V_4 = 4\times 12 = 48\ V.
Check: Sum of individual voltages equals total voltage: 12 + 20 + 40 + 48 = 120\ V.
Cost of Electrical Energy and Billing Concepts
Energy Charge: cost for electrical energy consumed; depends on usage (kWh).
Maximum Demand: the highest rate of energy consumption (in kW) during a billing period.
Demand Charge: a billing fee related to maximum demand.
Sample Problems: Energy and Billing Calculations
Problem 1: A 60 W lamp remains lit 24 hours/day for 30 days.
(a) Energy consumed: W = P t = 60\,\text{W} \times (24\ \text{h/day} \times 30\ \text{days}) = 60 \times 720 = 43{,}200\ \text{Wh} = 43.2\ \text{kWh}.
(b) Energy charge at Php 5.93 per kWh: \text{Charge} = 43.2\ \text{kWh} \times 5.93\ \text{Php/kWh} = \text{Php } 256.176.
Problem 2: A residence uses 1155 kWh in a billing period with a service charge of Php 261.18. Rates: first 1000 kWh at Php 5.92/kWh; next 2000 kWh at Php 5.52/kWh.
Energy charge: \text{Energy charge} = 261.18 + 5.92\times 1000 + 5.52\times 155 = 261.18 + 5{,}920 + 855.60 = \text{Php } 7{,}036.78.
Power and Energy: Worked Examples
Power is the rate of energy use; units: Watts (W).
Example: A bulb requiring 40 W for 3 hours consumes energy:
E = P t = 40\,\text{W} \times 3\,\text{h} \times 3600\,\text{s/h} = 432{,}000\ \text{J} = 432\ \text{kJ}.
Calculating Energy Bills: kWh Concept
Energy in kilowatt-hours (kWh) is used by energy suppliers to simplify costing.
Example: A 2000 W heater used for 5 hours:
Power in kW: 2\ \text{kW}.
Energy: E=\text{Power} \times \text{Time} = 2\ \text{kW} \times 5\ \text{h} = 10\ \text{kWh}.
Cost example (rate given as per kWh): if rate is Php 0.12 per kWh, cost = 10 \times 0.12 = Php 1.20. (Note: the transcript shows a rate example of 12 pence per kWh for a different currency context; adapt to your local currency.)
Notes on Units and Conversions
Power, energy, and billing rely on consistent use of SI units and common electrical units (W, V, A, Ω, J, s, and kWh).
When converting between units (e.g., W to kW, J to Wh), apply the appropriate conversion factors (1 kW = 1000 W, 1 kWh = 1000 Wh, 1 Wh = 3600 J).
Quick Reference Formulas (まとめ)
Resistance: R = \frac{\rho L}{A}.
Length dependence: \frac{R1}{R2}=\frac{L1}{L2}.
Area dependence: R \propto \frac{1}{A}.
Volume control as a resistor (conceptual): resistance changes with effective path length.
Ohm's Law: V = IR.
Power: P = VI = I^2R = \frac{V^2}{R}.
Energy: E = Pt.
Energy bills (example structure): base service charge + tiered energy charge across kWh usage.
Series circuit: RT = \sumi Ri,\quad VT = \sumi Vi,\quad IT = \frac{VT}{R_T}.
Sample numerical checks (series): verify voltages sum to applied voltage and that I is constant through all components.