Sciences · Topic 2.3
Physics — Electricity and Circuits
Electric circuits underpin modern technology. At Year 4 Advanced, you must not only apply Ohm's Law and circuit rules but justify circuit design choices, evaluate experimental methods, and link electrical concepts to real-world engineering applications.
What You'll Learn
- Define current (I), voltage (V), and resistance (R) with correct units
- Apply Ohm's Law (V = IR) to calculate unknown circuit quantities
- Calculate total resistance in series and parallel circuits
- Explain current and voltage behaviour in series vs parallel circuits
- Calculate electrical power (P = IV = V²/R = I²R) and energy (E = Pt)
- Evaluate the design of circuit experiments for validity and reliability
IB Assessment Focus
Criterion A: Apply circuit laws to complex multi-component circuits; solve for unknown quantities using appropriate formulae.
Criterion B: Design a valid investigation of Ohm's Law; justify the control of variables and choice of measurement technique.
Criterion C: Use correct SI units (A, V, Ω, W, J) consistently; present data in labelled tables and I-V graphs.
Criterion D: Evaluate experimental results for anomalies and uncertainty; discuss limitations of Ohm's Law model for real components.
Key Vocabulary
| Term | Definition |
|---|---|
| Current (I) | Flow of electric charge past a point; measured in amperes (A) |
| Voltage (V) | The potential difference (energy per unit charge) driving current; measured in volts (V) |
| Resistance (R) | Opposition to current flow; measured in ohms (Ω) |
| Ohm's Law | V = IR; applies to ohmic conductors at constant temperature |
| Series circuit | Components connected in a single loop; same current throughout |
| Parallel circuit | Components in separate branches; same voltage across each branch |
| Power (P) | Rate of energy transfer; P = IV; measured in watts (W) |
| EMF | Electromotive force — the energy supplied per unit charge by a source (battery/cell) |
Sciences · Topic 2.3
Current, Voltage, and Resistance
These three quantities are the fundamental variables of any electric circuit. Understanding their physical meaning — not just their formulas — is essential for justified analysis at Year 4.
Physical Meaning
| Quantity | Symbol | Unit | Physical meaning | Measured with |
|---|---|---|---|---|
| Current | I | Amperes (A) | Number of charges passing a point per second; rate of flow of charge (Q = It) | Ammeter (in series) |
| Voltage | V | Volts (V) | Energy given to (or transferred from) each coulomb of charge; "electrical pressure" | Voltmeter (in parallel) |
| Resistance | R | Ohms (Ω) | Opposition to current flow; caused by collisions between electrons and ions in the conductor | Calculated: R = V/I; or ohmmeter |
Why resistance increases with temperature (for metals):
As temperature rises, metal ions in the lattice vibrate more energetically. Free electrons collide more frequently with these vibrating ions, increasing opposition to flow. Resistance increases. This is why a filament bulb (tungsten wire at ~2500°C) has much higher resistance when hot than when cold — it is not ohmic over a temperature range.
As temperature rises, metal ions in the lattice vibrate more energetically. Free electrons collide more frequently with these vibrating ions, increasing opposition to flow. Resistance increases. This is why a filament bulb (tungsten wire at ~2500°C) has much higher resistance when hot than when cold — it is not ohmic over a temperature range.
Critical Rule: An ammeter must be connected in series (to measure the same current flowing through the component). A voltmeter must be connected in parallel (to measure the potential difference across the component). Reversing this will give incorrect readings and may damage instruments.
Sciences · Topic 2.3
Ohm's Law
Ohm's Law states that for an ohmic conductor at constant temperature, voltage is directly proportional to current. At Year 4, you must distinguish ohmic from non-ohmic behaviour and evaluate the limitations of the model.
Ohm's Law
V = IR I = V / R R = V / I
Ohmic vs Non-Ohmic Conductors
| Type | Example | I-V graph | Characteristic |
|---|---|---|---|
| Ohmic | Metal wire (constant T) | Straight line through origin | Resistance is constant; V ∝ I |
| Non-ohmic: bulb | Filament bulb | Curved — gradient decreases | Resistance increases as temperature increases |
| Non-ohmic: diode | Semiconductor diode | Flat then steep; no reverse current | Allows current in one direction only; very low resistance when forward-biased |
Example: A 12 V battery drives a current of 3 A through a resistor. Calculate the resistance.
- V = 12 V, I = 3 A.
- R = V / I = 12 / 3 = 4 Ω.
Limitation of Ohm's Law:
Ohm's Law is a model valid only for ohmic conductors at constant temperature. It breaks down for: (1) Components that heat up significantly (filament bulbs — resistance increases with temperature); (2) Semiconductor devices (diodes, transistors — non-linear relationship); (3) Electrolytes (ions, not electrons, carry charge — different mechanism). Acknowledging these limitations is required for Criterion D at Year 4.
Ohm's Law is a model valid only for ohmic conductors at constant temperature. It breaks down for: (1) Components that heat up significantly (filament bulbs — resistance increases with temperature); (2) Semiconductor devices (diodes, transistors — non-linear relationship); (3) Electrolytes (ions, not electrons, carry charge — different mechanism). Acknowledging these limitations is required for Criterion D at Year 4.
Sciences · Topic 2.3
Series and Parallel Circuits
The arrangement of components determines how current and voltage are distributed. Being able to justify WHY the rules work — not just state them — is the mark of Year 4 Advanced understanding.
Series Circuits
| Property | Rule | Justification |
|---|---|---|
| Current | Same throughout: I = I&sub1; = I&sub2; = ... | There is only one path; charge cannot accumulate, so the same number of electrons pass every point per second |
| Voltage | Divides: V = V&sub1; + V&sub2; + ... | Total energy per charge supplied by the battery is shared among the components |
| Total resistance | Rtotal = R&sub1; + R&sub2; + ... | Adding more components increases the total opposition to current flow |
Parallel Circuits
| Property | Rule | Justification |
|---|---|---|
| Current | Splits: I = I&sub1; + I&sub2; + ... | Current divides between branches; more charge takes each available path |
| Voltage | Same across each branch: V = V&sub1; = V&sub2; = ... | Each branch connects directly between the same two points (battery terminals) |
| Total resistance | 1/Rtotal = 1/R&sub1; + 1/R&sub2; + ... | Adding parallel paths provides more routes for current; total resistance decreases |
Why household wiring uses parallel circuits:
In a parallel circuit, each appliance receives the full supply voltage and operates independently. If one fails, the others continue to function (each has its own path). In a series circuit, one break or failure stops all current — all devices fail. This is why domestic electrical installations are wired in parallel.
In a parallel circuit, each appliance receives the full supply voltage and operates independently. If one fails, the others continue to function (each has its own path). In a series circuit, one break or failure stops all current — all devices fail. This is why domestic electrical installations are wired in parallel.
Critical Rule: For two resistors in parallel, total resistance is always less than the smallest individual resistance. This is because adding a parallel path always provides more routes for current, reducing the total opposition. If you calculate Rtotal > either R&sub1; or R&sub2;, you have made an error.
Sciences · Topic 2.3
Electrical Power and Energy
Power measures how quickly energy is transferred by an electrical component. Understanding and applying power equations allows you to evaluate the efficiency and cost of electrical devices.
Power Formulae
P = IV P = I²R P = V² / R
Energy and Cost
E = Pt (E in joules J, P in watts W, t in seconds s)
Deriving the Power Formulae
Starting from P = IV:
Substituting Ohm's Law (V = IR): P = I × IR = I²R.
Substituting I = V/R: P = (V/R) × V = V²/R.
All three formulae are derived from P = IV; knowing which to use depends on which quantities are known.
Substituting Ohm's Law (V = IR): P = I × IR = I²R.
Substituting I = V/R: P = (V/R) × V = V²/R.
All three formulae are derived from P = IV; knowing which to use depends on which quantities are known.
| Known quantities | Best formula | Example |
|---|---|---|
| I and V | P = IV | I = 2A, V = 12V → P = 24 W |
| I and R | P = I²R | I = 3A, R = 4Ω → P = 36 W |
| V and R | P = V²/R | V = 6V, R = 9Ω → P = 4 W |
Criterion D Tip: When evaluating an electrical experiment, always consider whether the component heated up during testing. If a resistor or bulb heats up, its resistance changes — any assumption that R is constant will introduce systematic error. State this as a limitation and suggest using low voltages or short measurement periods to minimise heating effects.
Sciences · Topic 2.3
Worked Examples
Multi-step solutions with full justification at Year 4 Advanced standard.
EXAMPLE 1A 12 V battery is connected to two resistors in series: R&sub1; = 4 Ω, R&sub2; = 8 Ω. Find (a) total resistance, (b) current, (c) voltage across each resistor.
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Full Solution
(a) Total resistance:
Rtotal = R&sub1; + R&sub2; = 4 + 8 = 12 Ω.
(b) Current:
In a series circuit, current is the same throughout.
I = V / Rtotal = 12 / 12 = 1 A.
(c) Voltage across each resistor:
V&sub1; = IR&sub1; = 1 × 4 = 4 V.
V&sub2; = IR&sub2; = 1 × 8 = 8 V.
Check: V&sub1; + V&sub2; = 4 + 8 = 12 V = supply voltage ✓ The voltage divides in proportion to resistance in a series circuit.
Rtotal = R&sub1; + R&sub2; = 4 + 8 = 12 Ω.
(b) Current:
In a series circuit, current is the same throughout.
I = V / Rtotal = 12 / 12 = 1 A.
(c) Voltage across each resistor:
V&sub1; = IR&sub1; = 1 × 4 = 4 V.
V&sub2; = IR&sub2; = 1 × 8 = 8 V.
Check: V&sub1; + V&sub2; = 4 + 8 = 12 V = supply voltage ✓ The voltage divides in proportion to resistance in a series circuit.
EXAMPLE 2Two resistors in parallel: R&sub1; = 6 Ω, R&sub2; = 3 Ω. Connected to a 12 V supply. Find (a) Rtotal, (b) total current, (c) current through each branch.
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Full Solution
(a) Total resistance:
1/Rtotal = 1/R&sub1; + 1/R&sub2; = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2.
Rtotal = 2 Ω. (Less than either individual resistance ✓)
(b) Total current:
Itotal = V / Rtotal = 12 / 2 = 6 A.
(c) Branch currents:
Same voltage (12 V) across each branch in a parallel circuit.
I&sub1; = V/R&sub1; = 12/6 = 2 A.
I&sub2; = V/R&sub2; = 12/3 = 4 A.
Check: I&sub1; + I&sub2; = 2 + 4 = 6 A = Itotal ✓ Currents add in a parallel circuit.
1/Rtotal = 1/R&sub1; + 1/R&sub2; = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2.
Rtotal = 2 Ω. (Less than either individual resistance ✓)
(b) Total current:
Itotal = V / Rtotal = 12 / 2 = 6 A.
(c) Branch currents:
Same voltage (12 V) across each branch in a parallel circuit.
I&sub1; = V/R&sub1; = 12/6 = 2 A.
I&sub2; = V/R&sub2; = 12/3 = 4 A.
Check: I&sub1; + I&sub2; = 2 + 4 = 6 A = Itotal ✓ Currents add in a parallel circuit.
EXAMPLE 3A kettle rated 2400 W is connected to a 240 V supply. Calculate (a) the current drawn, (b) the resistance of the heating element, (c) the energy used in 3 minutes.
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Full Solution
(a) Current:
P = IV → I = P / V = 2400 / 240 = 10 A.
(b) Resistance:
R = V / I = 240 / 10 = 24 Ω. (Or: R = V²/P = 240²/2400 = 57600/2400 = 24 Ω ✓)
(c) Energy in 3 minutes:
t = 3 × 60 = 180 s.
E = Pt = 2400 × 180 = 432 000 J = 432 kJ.
P = IV → I = P / V = 2400 / 240 = 10 A.
(b) Resistance:
R = V / I = 240 / 10 = 24 Ω. (Or: R = V²/P = 240²/2400 = 57600/2400 = 24 Ω ✓)
(c) Energy in 3 minutes:
t = 3 × 60 = 180 s.
E = Pt = 2400 × 180 = 432 000 J = 432 kJ.
EXAMPLE 4Explain why a bulb in a series circuit with another bulb is dimmer than if connected alone to the same battery.
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Full Solution
When two bulbs are in series, the total resistance of the circuit increases (Rtotal = R&sub1; + R&sub2;). By Ohm's Law (I = V/R), with higher total resistance and the same supply voltage, the current in the circuit decreases.
Since P = I²R, the power dissipated by each bulb is less (lower I). Less power means less light output — the bulbs are dimmer.
Additionally, in a series circuit, the voltage divides across both bulbs. Each bulb now has a smaller voltage across it than when alone — confirming lower power output (P = V²/R, with smaller V).
Contrast with parallel: If the bulbs were in parallel, each would receive the full supply voltage. Each would draw the same current as if alone and shine at full brightness, because they have separate paths back to the battery.
Since P = I²R, the power dissipated by each bulb is less (lower I). Less power means less light output — the bulbs are dimmer.
Additionally, in a series circuit, the voltage divides across both bulbs. Each bulb now has a smaller voltage across it than when alone — confirming lower power output (P = V²/R, with smaller V).
Contrast with parallel: If the bulbs were in parallel, each would receive the full supply voltage. Each would draw the same current as if alone and shine at full brightness, because they have separate paths back to the battery.
EXAMPLE 5Describe how you would investigate Ohm's Law using a wire, battery, ammeter, voltmeter, and variable resistor. Evaluate the design.
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Full Solution
Setup: Connect the wire (fixed resistor under test) in series with a variable resistor (rheostat) and ammeter. Connect a voltmeter in parallel across the wire. Connect to a battery.
Method: Adjust the rheostat to change the current through the circuit. At each setting, record I (from ammeter) and V (from voltmeter) across the wire. Repeat for 8–10 different current values.
Variables: Independent = voltage (adjusted via rheostat). Dependent = current. Controlled = type of wire (same material, length, diameter), temperature.
Analysis: Plot I-V graph. If Ohm's Law holds, it should be a straight line through the origin; gradient = 1/R.
Evaluation:
Strength: using a rheostat allows continuous variation of voltage, giving more data points and a more reliable line of best fit.
Limitation 1: The wire may heat up at high currents, increasing resistance and producing a non-linear graph — this would be a systematic error making the wire appear non-ohmic when it is. Solution: use low voltages and allow cooling time between readings.
Limitation 2: Internal resistance of the battery may affect readings at high currents. Solution: use an external voltage source with negligible internal resistance.
Method: Adjust the rheostat to change the current through the circuit. At each setting, record I (from ammeter) and V (from voltmeter) across the wire. Repeat for 8–10 different current values.
Variables: Independent = voltage (adjusted via rheostat). Dependent = current. Controlled = type of wire (same material, length, diameter), temperature.
Analysis: Plot I-V graph. If Ohm's Law holds, it should be a straight line through the origin; gradient = 1/R.
Evaluation:
Strength: using a rheostat allows continuous variation of voltage, giving more data points and a more reliable line of best fit.
Limitation 1: The wire may heat up at high currents, increasing resistance and producing a non-linear graph — this would be a systematic error making the wire appear non-ohmic when it is. Solution: use low voltages and allow cooling time between readings.
Limitation 2: Internal resistance of the battery may affect readings at high currents. Solution: use an external voltage source with negligible internal resistance.
EXAMPLE 6A fuse is rated 3 A. A 690 W hairdryer is connected to a 230 V supply. Will the fuse blow?
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Full Solution
P = IV → I = P/V = 690/230 = 3 A.
The hairdryer draws exactly 3 A. The fuse is rated 3 A, meaning it will blow at currents exceeding 3 A. In practice, a fuse is designed to protect against fault currents above its rating, not normal operating current that equals its rating.
At normal operation, the current equals the fuse rating exactly — the fuse will not blow under normal use. However, this provides minimal protection: if the hairdryer developed even a small fault drawing additional current (e.g., 3.1 A), the fuse would blow and protect the circuit.
Recommendation: A 3 A fuse is appropriate for this appliance (operating current = 3 A, which is at the limit). A 5 A fuse would allow fault currents up to 5 A before blowing, offering less protection. The 3 A fuse is the correct choice.
The hairdryer draws exactly 3 A. The fuse is rated 3 A, meaning it will blow at currents exceeding 3 A. In practice, a fuse is designed to protect against fault currents above its rating, not normal operating current that equals its rating.
At normal operation, the current equals the fuse rating exactly — the fuse will not blow under normal use. However, this provides minimal protection: if the hairdryer developed even a small fault drawing additional current (e.g., 3.1 A), the fuse would blow and protect the circuit.
Recommendation: A 3 A fuse is appropriate for this appliance (operating current = 3 A, which is at the limit). A 5 A fuse would allow fault currents up to 5 A before blowing, offering less protection. The 3 A fuse is the correct choice.
Sciences · Topic 2.3
Practice Q&A
Attempt each question before revealing the model answer. Always state the formula before substituting values.
CALCULATEA 12 V battery drives a current of 3 A through a resistor. Calculate the resistance.
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Model Answer
R = V / I = 12 / 3 = 4 Ω.
CALCULATEThree resistors in series: 2 Ω, 5 Ω, 8 Ω. What is the total resistance?
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Model Answer
Rtotal = R&sub1; + R&sub2; + R&sub3; = 2 + 5 + 8 = 15 Ω.
EXPLAINWhy is total resistance less than either individual resistor in a parallel circuit?
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Model Answer
Adding a parallel branch provides an additional path for current to flow. More available paths means there is less total opposition to current flow overall. The total resistance formula 1/Rtotal = 1/R&sub1; + 1/R&sub2; always gives Rtotal less than either component. For example, two 10 Ω resistors in parallel give Rtotal = 5 Ω — less than either alone.
CALCULATEA 60 W bulb is connected to a 240 V supply. Find the current and resistance of the filament.
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Model Answer
P = IV → I = P/V = 60/240 = 0.25 A.
R = V/I = 240/0.25 = 960 Ω. (Alternatively: R = V²/P = 57600/60 = 960 Ω ✓)
R = V/I = 240/0.25 = 960 Ω. (Alternatively: R = V²/P = 57600/60 = 960 Ω ✓)
EVALUATEA student's I-V graph for a filament bulb curves away from the origin. Explain why and what it tells us about the bulb.
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Model Answer
As current through the filament increases, the filament's temperature increases significantly (it glows). This causes the tungsten ions in the lattice to vibrate more energetically, increasing the rate of electron-ion collisions and therefore the resistance. As resistance increases, the current increases more slowly than voltage (the gradient decreases). The bulb is non-ohmic: its resistance is not constant. Ohm's Law does not apply because the conditions (constant temperature) are not met.
CALCULATER&sub1; = 4 Ω and R&sub2; = 12 Ω in parallel. Find Rtotal.
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Model Answer
1/Rtotal = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3.
Rtotal = 3 Ω. (Less than 4 Ω ✓)
Rtotal = 3 Ω. (Less than 4 Ω ✓)
DISCUSSEvaluate the limitations of Ohm's Law as a model.
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Model Answer
Ohm's Law (V = IR with constant R) is a useful model but has significant limitations:
1. It only applies to ohmic conductors at constant temperature. For real metal wires, resistance increases with temperature — making them non-ohmic if temperature changes significantly.
2. Non-ohmic devices (filament bulbs, diodes, thermistors, LDRs) do not obey Ohm's Law at all — their resistance changes with current, temperature, or light intensity.
3. For very high voltages, even materials that appear ohmic may break down (dielectric breakdown).
Despite these limitations, Ohm's Law is an excellent approximation for many practical resistors at normal operating temperatures and is the foundation of circuit analysis. Recognising its scope of validity is the mark of sophisticated physical understanding.
1. It only applies to ohmic conductors at constant temperature. For real metal wires, resistance increases with temperature — making them non-ohmic if temperature changes significantly.
2. Non-ohmic devices (filament bulbs, diodes, thermistors, LDRs) do not obey Ohm's Law at all — their resistance changes with current, temperature, or light intensity.
3. For very high voltages, even materials that appear ohmic may break down (dielectric breakdown).
Despite these limitations, Ohm's Law is an excellent approximation for many practical resistors at normal operating temperatures and is the foundation of circuit analysis. Recognising its scope of validity is the mark of sophisticated physical understanding.
APPLYAn electric heater draws 5 A at 240 V for 2 hours. Calculate the energy used in joules and the cost at £0.30 per kWh.
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Model Answer
P = IV = 5 × 240 = 1200 W = 1.2 kW.
t = 2 hours = 2 × 3600 = 7200 s.
E = Pt = 1200 × 7200 = 8 640 000 J = 8.64 MJ.
Energy in kWh = 1.2 kW × 2 h = 2.4 kWh.
Cost = 2.4 × £0.30 = £0.72.
t = 2 hours = 2 × 3600 = 7200 s.
E = Pt = 1200 × 7200 = 8 640 000 J = 8.64 MJ.
Energy in kWh = 1.2 kW × 2 h = 2.4 kWh.
Cost = 2.4 × £0.30 = £0.72.
Sciences · Review
Flashcard Review
Tap each card to reveal the answer. Try to answer from memory first.
State Ohm's Law and its three forms.
V = IR | I = V/R | R = V/I
Valid for ohmic conductors at constant temperature.
Valid for ohmic conductors at constant temperature.
Tap to reveal
Total resistance formula for a series circuit?
Rtotal = R&sub1; + R&sub2; + R&sub3; + ...
Adding components in series always increases total resistance.
Adding components in series always increases total resistance.
Tap to reveal
Total resistance formula for a parallel circuit?
1/Rtotal = 1/R&sub1; + 1/R&sub2; + ...
Total resistance is always less than the smallest individual resistance.
Total resistance is always less than the smallest individual resistance.
Tap to reveal
In a series circuit: what is the same throughout?
Current (I) is the same throughout a series circuit. Voltage divides across components in proportion to resistance.
Tap to reveal
In a parallel circuit: what is the same for each branch?
Voltage (V) is the same across each branch. Current splits between branches in inverse proportion to resistance.
Tap to reveal
State the three power formulae.
P = IV | P = I²R | P = V²/R
All derived from P = IV by substituting Ohm's Law.
All derived from P = IV by substituting Ohm's Law.
Tap to reveal
How do you measure current and voltage in a circuit?
Current (I): ammeter connected IN SERIES.
Voltage (V): voltmeter connected IN PARALLEL across the component.
Voltage (V): voltmeter connected IN PARALLEL across the component.
Tap to reveal
What is an ohmic conductor?
A conductor where V ∝ I (constant resistance) at constant temperature. The I-V graph is a straight line through the origin. Example: metal wire (not a bulb).
Tap to reveal
Why do houses use parallel circuits?
Each appliance receives full supply voltage; appliances operate independently; if one fails, the rest continue to work. Series circuits would fail completely if one component breaks.
Tap to reveal
Why does resistance in a metal wire increase with temperature?
Higher temperature causes metal ions to vibrate more energetically. Free electrons collide more frequently with vibrating ions, increasing opposition to current flow.
Tap to reveal
Energy formula for electrical devices?
E = Pt (joules = watts × seconds)
Also: E = VIt = I²Rt
Also: E = VIt = I²Rt
Tap to reveal
What is the unit of electrical power?
Watts (W). 1 W = 1 J/s = 1 V × A. For appliances, often expressed in kilowatts (kW): 1 kW = 1000 W.
Tap to reveal
Limitation of Ohm's Law?
Only valid for ohmic conductors at constant temperature. Fails for filament bulbs (resistance increases with temperature), diodes, and thermistors — all have non-linear I-V relationships.
Tap to reveal
What is EMF?
Electromotive force — the energy supplied per unit charge by a source (e.g., battery). Measured in volts (V). Not the same as the terminal voltage (which is reduced by internal resistance).
Tap to reveal
What happens to total resistance when you add more resistors in parallel?
Total resistance decreases (more paths for current). It will always be less than the smallest individual resistor in the parallel combination.
Tap to reveal
Sciences · Practice Test
Practice Test — 20 Questions
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