Electricity Short Answer
| # | Answer |
|---|---|
| 1 | Electric current is the rate of flow of electric charge through a point in a circuit. Its SI unit is the ampere (A). |
| 2 | Ohm’s law states that the current through a conductor is directly proportional to the potential difference across it, provided temperature remains constant. The quantities are voltage, current, and resistance. |
| 3 | Using V = IR, V = 2.5 × 4.0 = 10 V. |
| 4 | For a metal conductor, resistance generally increases as temperature increases. This is because the ions vibrate more, causing more collisions for the moving electrons. |
| 5 | Electric potential difference is the work done per unit charge to move charge between two points in a circuit. It is measured in volts (V). |
| 6 | Using I = Q/t, I = 12/3 = 4 A. |
| 7 | In a series circuit, components are connected in one path so the same current flows through all of them. In a parallel circuit, components are connected in separate branches so the voltage across each branch is the same. |
| 8 | For series resistors, resistances add: Rtotal = 6 + 3 = 9 Ω. |
| 9 | Using 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R = 2 Ω. |
| 10 | House wiring is connected in parallel so each appliance gets the full supply voltage and can operate independently. If one device is turned off or fails, the others still work. |
| 11 | First find current: I = V/R = 12/4 = 3 A. Then power is P = VI = 12 × 3 = 36 W. |
| 12 | Electrical energy transferred is given by E = VIt. Here E is energy in joules, V is potential difference in volts, I is current in amperes, and t is time in seconds. |
| 13 | Use E = VIt. Time = 10 min = 600 s, so E = 240 × 5 × 600 = 720000 J. |
| 14 | A conductor allows charge to flow easily, an insulator strongly resists charge flow, and a semiconductor has conductivity between the two and can be controlled by conditions such as temperature or doping. |
| 15 | Electromotive force is the energy supplied by a source per unit charge. It is the total energy provided to charges by the source, measured in volts. |
| 16 | Terminal voltage V = emf − Ir = 1.5 − (1.2 × 0.5) = 1.5 − 0.6 = 0.9 V. |
| 17 | When current increases, the voltage drop across the internal resistance increases because V = Ir. This leaves less voltage available at the terminals. |
| 18 | Current in 2 Ω branch: I = 12/2 = 6 A. Current in 4 Ω branch: I = 12/4 = 3 A. Total current = 6 + 3 = 9 A. |
| 19 | Kirchhoff’s current law states that the total current entering a junction equals the total current leaving it. Kirchhoff’s voltage law states that the sum of potential differences around a closed loop is zero. |
| 20 | Total resistance in series = 5 + 7 = 12 Ω. Current is I = V/R = 18/12 = 1.5 A. |
| 21 | Using R = V²/P, the 30 W lamp has the larger resistance because lower power at the same voltage means higher resistance. Specifically, R30 = 120²/30 is greater than R60 = 120²/60. |
| 22 | A fuse protects a circuit by melting when the current becomes too large. This breaks the circuit and prevents overheating or damage. |
| 23 | 100 W = 0.1 kW, so energy = 0.1 × 4 = 0.4 kWh. |
| 24 | Resistance is proportional to length and inversely proportional to area. Doubling the length and halving the area makes the resistance 4 times larger. |
| 25 | No, it is not always true. According to Ohm’s law, V = IR, so voltage depends on both current and resistance; a larger current only means a larger voltage if resistance stays constant. |