Voltage in a transformer is determined by the turns ratio between the primary and secondary windings. The secondary voltage equals the primary voltage times the secondary turns divided by the primary turns. Here, there are 1000 turns on the primary and 500 on the secondary, so the ratio Ns/Np is 500/1000 = 0.5. With an input of 120 V, the output becomes 120 V × 0.5 = 60 V. This is a step-down transformer, since the secondary has fewer turns than the primary, reducing the voltage. If you think about power, the current on the secondary would rise to roughly conserve power (ignoring losses), but the question asks for voltage, so the result is 60 V. The other voltages would require different turns relationships (for example, equal turns give 120 V, more turns on the secondary would give voltages above 120 V, and so on). The correct output is 60 V.

Silicon is used as a semiconductor because its electronic structure and fabrication properties let its conductivity be precisely controlled. It has a crystal lattice made from four-valence electrons, which forms a stable covalent network with a characteristic energy band gap of about 1.12 eV. This band gap is large enough to keep it as a good insulator at absolute zero but small enough to allow controlled conduction at room temperature when doped. By introducing tiny amounts of dopants, silicon can become n-type (extra electrons) or p-type (holes), creating regions with different charge carriers that form essential devices like diodes and transistors. Its ability to form a high-quality native oxide, silicon dioxide, provides excellent insulation and a convenient surface for layering electronic structures during manufacturing. Silicon is also abundant and yields uniform, high-purity wafers ideal for large-scale integrated circuits. In contrast, copper is a metal with plenty of free electrons and serves as a conductor, not a semiconductor. Iron is also a metal. Plastics are typically insulators, though some specialty polymers can be engineered for other roles, they are not the standard semiconductor material used for most electronics.

Discharging a capacitor safely means giving it a controlled path to release its stored energy without creating sparks or shock. Placing a resistor directly across the terminals provides that controlled bleed. The high resistance limits the discharge current, so energy is released gradually rather than dumped all at once. Shorting the terminals with a metal tool creates an almost direct short circuit. The capacitor would try to dump its energy instantly, causing a large surge current, arcing, heat, possible damage to the capacitor or surrounding components, and a real risk of shock. It’s unpredictable and dangerous. Connecting it to a power source would keep it charged or even re-energize it, not discharge it. Letting it sit unplugged doesn’t guarantee discharge either; many capacitors retain a charge for a long time, and you could still be shocked when handling them. Using a resistor of about 15,000 to 20,000 ohms across the terminals provides a safe discharge path. The current is limited to a few milliamps, and the resistor dissipates the energy as heat in a predictable way (P = V^2 / R). With this setup, you’ll discharge it gradually to a safe voltage while keeping risk low. The wattage rating (2–5 W) ensures the resistor can safely handle the heat produced during discharge.

Power in an electrical circuit is found by multiplying voltage by current: P = V × I. With a 120-volt supply delivering 2 amperes, the power is 120 × 2 = 240 watts. This shows how the load converts electrical energy into useful power. For comparison, using the same voltage, a smaller current (like 1 A) would give 120 W, while a larger current (like 4 A) would give 480 W, and 0.5 A would give 60 W.