The powers of two light-bulbs are 60 w and 120 w when connected in parallel across a 110-v source. what is the power of the bulb that will output less power when the two light-bulbs are connected in series with the same source?

The power of the bulbs is : P1 = 26.67 W and P2 = 13.66 W

Explanation:

Step 1:  Calculate the resistance R of each light bulb

P = U*I = U²/R

R1 = U²/P1 = 110²/60 = 201.67 Ω

R2 = U²/P2 = 110²/120 = 100.83 Ω

Serial

Step 2: Calculate the current through the bulbs

I = U /( R1 + R2)

I = 110 V/ (201.67 Ω + 100.83Ω)

I = 0.3636 A

Step 3: Calculate the power of each bulb

P = U*I = I²*R

P1 = (0.3636)² * 201.67 = 26.67 W

P2 = (0.3636)² *101.83 = 13.66 W

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

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