Final answer:
To reduce an oven's current from 16 A to 12 A at 220 V, a 5.42 Ω resistor must be added in series, resulting in a voltage of 65.04 V across this resistor. This demonstrates the application of Ohm's Law in managing electric power.
Step-by-step explanation:
To reduce current to 12 A in an oven that takes 16 A at 220 V, we must first calculate the original resistance of the oven using Ohm's Law (V = IR), which gives us R = V / I. For the oven, the resistance (Roven) is 220 V / 16 A = 13.75 Ω. To reduce the current to 12 A, the total current that flows through the oven and the additional resistor in series must be 12 A.
We use the formula for total resistance in series: Rtotal = Roven + Rnew. The voltage across the oven and new resistor remains 220 V, so we have V = Itotal × Rtotal. Substituting the values, we get 220 V = 12 A × (13.75 Ω + Rnew). Solving for Rnew, the resistance to be added in series comes to:
Rnew = (220 V / 12 A) - 13.75 Ω = 5.42 Ω.
To find the voltage across this new resistor, use Ohm's Law (V = IR), which gives us Vresistor = I × Rnew = 12 A × 5.42 Ω = 65.04 V. Therefore, the resistance to be connected in series is 5.42 Ω, and the voltage across this resistor is 65.04 V.
Knowing how to calculate resistance and voltage is crucial for safely managing electric power in household appliances and industrial equipment.