🔌 بنك الأسئلة: التيار الكهربائي والدوائر الكهربائية

💡 Concept: A resistor is an electrical component that opposes the flow of current and converts electrical energy into heat energy (Joule heating).

📌 Analysis: • Battery: converts chemical energy to electrical energy.
• Resistor: converts electrical energy to heat only.
• Lamp: converts electrical energy to light and heat.
• Hair dryer: converts electrical energy to heat and kinetic energy (fan).

💡 Tip: A resistor is the simplest device that converts electrical energy completely into heat.

💡 Concept: In a circuit, energy is lost as heat in components that have resistance. The resistor converts electrical energy into heat (Joule heating).

📌 Analysis: • Ammeter: has very low resistance, negligible energy loss.
• Voltmeter: has very high resistance, negligible current flow.
• Resistor: has significant resistance, converts electrical energy to heat.
• Wires: have very low resistance, negligible energy loss.

💡 Tip: In a circuit, energy is lost mainly in the resistor because \(P = I^2R\).

💡 Concept: Electrical power is \(P = VI\), and energy is \(E = P \times t = VIt\).

📌 Calculate power: \(P = VI = 20 \times 3 = 60 \;\;W\)

📌 Convert time: \(t = 1 \;\;min = 60 \;\;s\)

📌 Calculate energy: \(E = P \times t = 60 \times 60 = 3600 \;\;J\)

💡 Tip: Always convert time to seconds when using SI units (Joules).

📌 Total charge: \(Q=ne=20\times10^{20}\times1.6\times10^{-19}=320\,C\).

📌 \(I=\dfrac{Q}{t}=\dfrac{320}{120}\approx2.7\,A\).

💡 Concept: Power is the rate of doing work: \(P = \frac{W}{t} = \frac{F \times d}{t} = \frac{kg \cdot m/s^2 \times m}{s} = \frac{kg \cdot m^2}{s^3}\).

📌 Derive unit: \(Watt = \frac{Joule}{second} = \frac{kg \cdot m^2/s^2}{s} = kg \cdot m^2 \cdot s^{-3}\).

💡 Tip: 1 Watt = 1 Joule/second = 1 kg·m²/s³.

📌 Two components are in parallel when both their ends connect to the same two nodes, so they share the same potential difference.

📌 In the figure the lamp’s two terminals connect to the two ends of the variable resistor, so they are in parallel. The ammeter, fuse, battery and switch are in series, and so are \(R\) and the variable resistor.

📌 For an ohmic resistor \(R=\dfrac{V}{I}\) (the V–I relation has a constant slope).

📌 From the graph: at \(V=8\,V\) we read \(I=0.4\,A\) (likewise \(4\,V\to0.2\,A\)).

📌 \(R=\dfrac{8}{0.4}=20\,\Omega\).

💡 Concept: For the same voltage, a larger current means smaller resistance. The slope of the I-V graph is \(1/R\).

📌 Analyze slopes: • R1 and R3 have the same slope (same current for same voltage) → equal resistance.
• R2 has a steeper slope (larger current for same voltage) → smaller resistance.
Therefore: \(R_1 = R_3 > R_2\).

💡 Tip: Steeper slope in an I-V graph means lower resistance.

📌 The ohmic line is straight: \(R=\dfrac{8}{0.4}=20\,\Omega\), i.e. a constant I–V slope of \(0.05\,A/V\).

📌 The lamp’s curve flattens gradually (its resistance rises as the filament heats up), so its slope decreases with voltage.

📌 Between \(3\,V\) and \(4\,V\) the current changes from about \(0.37\,A\) to \(0.42\,A\): a slope of \(\approx0.05\,A/V\), equal to the ohmic line’s, i.e. at \(V\approx3.5\,V\).

💡 Concept: The resistance of the lamp is constant: \(R = \frac{V^2}{P}\). Then use \(P = \frac{V^2}{R}\) for the new voltage.

📌 Calculate resistance: \(R = \frac{V^2}{P} = \frac{220^2}{100} = 484 \;\Omega\)

📌 Calculate new power: \(P_{new} = \frac{V_{new}^2}{R} = \frac{50^2}{484} = \frac{2500}{484} = 5.17 \approx 5.2 \;\;W\)

💡 Tip: The resistance of a lamp is constant (assuming it doesn't change with temperature).

💡 Concept: Power is \(P = VI\), so \(I = \frac{P}{V}\).

📌 Calculate current: \(I = \frac{P}{V} = \frac{120}{220} = 0.545 \approx 0.54 \;\;A\)

💡 Tip: Remember the power triangle: \(P = VI\), \(I = P/V\), \(V = P/I\).

💡 Concept: Electrical energy is \(E = VIt\).

📌 Convert time: \(t = 1 \;\;min = 60 \;\;s\)

📌 Calculate energy: \(E = VIt = 220 \times 0.1 \times 60 = 1320 \;\;J\)

💡 Tip: Energy = Power × Time = Voltage × Current × Time.

💡 Concept: Resistance is given by \(R = \rho \frac{L}{A}\). For the same material (\(\rho\) constant), resistance is proportional to length and inversely proportional to area.

📌 Analyze each wire: Wire 3 has the largest cross-sectional area and the shortest length → least resistance.

💡 Tip: To minimize resistance: use short, thick wires with low resistivity.

📌 \(R=\rho\dfrac{L}{A}\): resistance increases with resistivity (aluminium is higher than copper, and it rises with temperature), with length, and as the cross-section decreases.

📌 Resistor C: aluminium at 80°C, long and very thin ⇒ the greatest resistance.

📌 Resistors A and B have the same dimensions and temperature, and aluminium (B) has higher resistivity than copper (A) ⇒ \(R_B>R_A\).

📌 Resistor D: copper, short and thick ⇒ the smallest resistance. So \(R_C>R_B>R_A>R_D\).

💡 Concept: When connected to the same battery (same voltage), the lamp with the smallest resistance will draw the most current and consume the most power: \(P = V^2/R\).

📌 Analyze: Lamp\(A\)has the lowest resistance (thin filament) → highest power → most energy consumed in one hour.

💡 Tip: Energy consumed = Power × Time. For the same time, higher power means more energy.

💡 Concept: To minimize power loss \(P = I^2R\) in transmission lines, resistance should be as low as possible: \(R = \rho L/A\).

📌 Requirements: • Low resistivity (\(\rho\) small)
• Large cross-sectional area (\(A\) large)
These two factors minimize resistance.

💡 Tip: Transmission lines use materials like copper or aluminium (low resistivity) with large diameters.

💡 Concept: To reduce power loss during transmission, high voltage and low current are used \(P = VI\). At the city, voltage is stepped down for safe use.

📌 Analysis: • At station: Step-up transformer (increase V, decrease I) to reduce \(I^2R\) losses.
• At city: Step-down transformer (decrease V, increase I) for safe consumption.

💡 Tip: Power loss in transmission lines is proportional to \(I^2\). Higher voltage means lower current for the same power.

📌 An ammeter has a very small resistance and is connected in series; the variable resistor and the fuse are also in series and carry the current.

📌 A voltmeter has a very large resistance (approaching infinity) and is connected in parallel; if connected in series it practically stops the current in the circuit.

📌 \(R=\rho\dfrac{L}{A}\) for the same material, length and temperature gives \(R\propto\dfrac1A\); since \(A_{R1}=2A_{R2}\), \(R_1=\dfrac{R_2}{2}\).

📌 Each resistor is connected to the same battery with the same stored energy \(E\): \(P=\dfrac{V^2}{R}\) and \(t=\dfrac{E}{P}=\dfrac{ER}{V^2}\), so \(t\propto R\).

📌 Resistor 2 has twice the resistance of resistor 1, so it runs twice as long: \(t_2=2t_1\).

💡 Concept: Efficiency = (useful energy) / (total energy). Heat energy = total energy - useful energy.

📌 Total energy in 1 hour: \(E_{total} = P \times t = 100 \times 3600 = 3.6 \times 10^5 \;\;J\)

📌 Useful energy (radiation): \(E_{useful} = 0.40 \times E_{total} = 0.40 \times 3.6 \times 10^5 = 1.44 \times 10^5 \;\;J\)

📌 Heat energy: \(E_{heat} = E_{total} - E_{useful} = 3.6 \times 10^5 - 1.44 \times 10^5 = 2.16 \times 10^5 \;\;J\)

💡 Tip: Heat energy = Total energy × (1 - Efficiency).

📌 Consumption = current reading − previous reading: \(236458-234532=1926\,kWh\).

📌 Cost = consumption × unit price: \(1926\times0.3=577.8\,DH\).

💡 Concept: Power \(P = VI KW\). Energy in \(E=P(kW) \times t(h)\) Cost = Energy × Price per kWh.

📌 Calculate power: \(P = VI = 220 \times 15 = 3300 \;\;W = 3.3 \;\;kW\)

📌 Total energy for the week: \(E = \frac{Cost}{Price} = \frac{60}{0.3} = 200 \;\;kWh\)

📌 Total time for the week: \(t_{week} = \frac{E}{P} = \frac{200}{3.3} = 60.61 \;\;h\)

📌 Time per day: \(t_{daily} = \frac{60.61}{7} = 8.66 \approx 8.65 \;\;h\)

💡 Tip: Remember to convert Watts to Kilowatts when using kWh units.

📌 Current is the rate of flow of charge \(I=\dfrac{Q}{t}\), so \(Q=It=5\times20=100\,C\).

📌 Ohm’s law: \(R=\dfrac{V}{I}=\dfrac{12}{3}=4\,\Omega\).

📌 \(R=\rho\dfrac{L}{A}\): resistance is directly proportional to length when material and area are fixed.

📌 Doubling the length doubles the resistance: \(2R\).

📌 Convert the power to kilowatts: \(40\,W=0.04\,kW\).

📌 \(E=Pt=0.04\times2=0.08\,kWh\).

📌 Electric power: \(P=VI\).

📌 \(P=220\times5=1100\,W\).

📌 \(A=\pi r^2\): doubling the radius multiplies the area by 4.

📌 \(R\propto\dfrac1A\) ⇒ \(R_2=\dfrac R4\).

📌 \(P=\dfrac{V^2}{R}\Rightarrow R=\dfrac{V^2}{P}\).

📌 \(R=\dfrac{120^2}{60}=240\,\Omega\).

📌 Power is proportional to the square of the voltage when resistance is constant: \(P\propto V^2\).

📌 The voltage is half the rated voltage, so the power is a quarter: \(P=\dfrac{100}{4}=25\,W\).

📌 Energy: \(E=Pt=2\times3=6\,kWh\).

📌 Cost: \(6\times0.3=1.8\,DH\).

📌 Total energy: \(E=3\times4\times30=360\,kWh\).

📌 The bill: \(360\times0.3=108\,DH\).

📌 Charge: \(Q=It=3.2\times10=32\,C\).

📌 Number of electrons: \(n=\dfrac{Q}{e}=\dfrac{32}{1.6\times10^{-19}}=2\times10^{20}\).

📌 \(\dfrac{R_2}{R_1}=\dfrac{L_2}{L_1}\times\dfrac{A_1}{A_2}\) since the material and temperature are the same.

📌 \(\dfrac{R_2}{R_1}=3\times\dfrac12=1.5\).

📌 As the voltage rises the current rises and the filament heats up, so its resistance increases (\(R=\dfrac{V}{I}\) grows).

📌 So the current is not proportional to the voltage and the curve flattens; the lamp is a non-ohmic conductor.

📌 Total energy: \(E=Pt=60\times3600=2.16\times10^{5}\,J\).

📌 Heat is \(80\%\) of the total: \(0.8\times2.16\times10^{5}\approx1.73\times10^{5}\,J\).

📌 \(P=I^2R\): the lost power is proportional to the square of the current when the resistance is constant.

📌 \(I\to\tfrac I2\Rightarrow P\to\dfrac{P}{4}\); that is why voltage is stepped up on transmission lines to reduce the current.

Solution: a) Electrons (negative charges) move from lower potential (B) to higher potential (A) because they are attracted to the higher potential. While conventional current (scientists' convention) moves from higher potential (A) to lower potential (B) (opposite to electron direction).

b) When the potentials are equal, the potential difference stops, and thus charge movement stops because the electric field inside the wire becomes zero.

c) The battery acts as a charge pump, maintaining a constant potential difference between its terminals by converting chemical energy into electrical energy, ensuring continuous charge movement.

Solution: 1. Potential energy (water stored in the dam) →

2. Kinetic energy (falling water) →

3. Mechanical energy (turbine rotation) →

4. Electrical energy (generator converts rotational motion to electric current via electromagnetic induction).

5. Electrical energy is transmitted through power lines to the home.

6. In the lamp, electrical energy is converted to light energy (and mostly to heat energy).

Solution: Factors: 1. Length (L): Resistance increases proportionally with length (R ∝ L).

2. Cross-sectional area (A): Resistance decreases inversely with area (R ∝ 1/A).

3. Material type (resistivity ρ): Resistivity varies with different materials (R = ρ L / A).

4. Temperature: In metals, resistance increases with temperature due to atomic vibrations and increased electron collisions.

Lamp explanation: When the lamp is turned on, the tungsten filament temperature rises significantly, increasing its resistance. Therefore, the lamp's resistance when operating is higher than when it is cold (off).

Solution: a) Ohm's law states: "The potential difference across a metal conductor is directly proportional to the current passing through it, provided temperature is constant and the conductor is ohmic." Mathematically: V = I × R.

b) To verify: Change the voltage and measure the current, then calculate the ratio V/I each time. If the ratio is constant and equals 100 Ω, it is ohmic.
The graph will be a straight line passing through the origin, with slope equal to the resistance value.

c) Example of non-ohmic resistor: Diode or fluorescent lamp. Reason: The relationship between V and I is not linear because resistance changes with voltage and current (thermal effects or semiconductor properties).

Solution: a) Current: I = P / V = 2000 / 220 = 9.09 A

b) Energy: P = 2000 W = 2 kW. Time = 1.5 h.
Energy (kWh) = 2 × 1.5 = 3 kWh.
Energy (J) = 3 × 3.6×10⁶ = 10.8×10⁶ J (or 10,800,000 J).

c) Daily consumption = 2 kW × 5 h = 10 kWh.
Monthly consumption = 10 × 30 = 300 kWh.
Cost = 300 × 0.30 = 90 dinars.

Solution: a) From resistance law: R = ρ L / A → A = ρ L / R
A = (1.6×10⁻⁸ × 5) / 0.5 = (8×10⁻⁸) / 0.5 = 1.6×10⁻⁷ m²

b) A = 2 mm² = 2×10⁻⁶ m².
L = R × A / ρ = (0.5 × 2×10⁻⁶) / (1.6×10⁻⁸) = (1×10⁻⁶) / (1.6×10⁻⁸) = 62.5 m

c) If resistivity doubles (ρ₂ = 2ρ₁) and we want the same resistance and same area, the length must also double (since R ∝ ρL). Therefore, the length becomes twice the length calculated in part (b).

Solution: a) Total current = Total power of devices / Voltage
Pₜ = 1500 + 100 + 1200 = 2800 W
Iₜ = 2800 / 220 = 12.73 A
Since \(15 A > 12.73A\) the breaker will not trip.

b) With the washing machine added: Pₜ = 2800 + 2000 = 4800 W
Iₜ = 4800 / 220 = 21.82 A
This current is greater than 15 A, so the breaker will trip (or the fuse will blow) to protect the wires from fire.
Solution: Either use a breaker with higher rating (e.g., 25 A) while ensuring wire capacity, or operate devices sequentially rather than simultaneously, or distribute loads on separate circuits.