A 6O-W, 120-V light bulb is to be connected in series with a resistor across a 277-V line. What is the resistanceand minimum power rating of the resistor required if the light bulb is to operate under rated conditions?

To solve this problem, we need to follow these steps:

Step 1: Find the current in the light bulb. The power P of the light bulb is given by P = VI, where V is the voltage and I is the current. We can rearrange this to find I = P/V. Substituting the given values, we get I = 60W / 120V = 0.5A.

Step 2: Find the voltage across the resistor. The light bulb and the resistor are in series, so the total voltage across them is the sum of their individual voltages. The total voltage is given as 277V, and the voltage across the light bulb is 120V, so the voltage across the resistor Vr is 277V - 120V = 157V.

Step 3: Find the resistance of the resistor. Ohm's law states that V = IR, where V is the voltage, I is the current, and R is the resistance. We can rearrange this to find R = V/I. Substituting the values we found, we get R = 157V / 0.5A = 314Ω.

Step 4: Find the minimum power rating of the resistor. The power P of the resistor is given by P = VI. Substituting the values we found, we get P = 157V * 0.5A = 78.5W.

So, the resistance of the resistor required is 314Ω and its minimum power rating is 78.5W.