(i) two 110v lamps are connected in parallel. their ratings are 150w and 200w.
determine the value of a resistor which when wired in series with the lamps will enable them to
operate from 230v supply.

37.7 Ω

Explanation:

The resistance in each lamp is:

P = IV = V²/R

R = V²/P

R₁ = (110 V)² / (150 W)

R₁ = 80.67 Ω

R₂ = (110 V)² / (200 W)

R₂ = 60.5 Ω

Their combined resistance in parallel is:

1/R = 1/R₁ + 1/R₂

1/R = 1/(80.67 Ω) + 1/(60.5 Ω)

R = 34.57 Ω

So the current that goes through them is:

V = IR

I = V/R

I = (110 V) / (34.57 Ω)

I = 3.182 A

We need the same current in the new circuit.

V = IR

230 V = (3.182 A) (R₃ + 34.57 Ω)

R₃ = 37.7 Ω

We can check our answer by finding the voltage drop across this resistor:

V = I R₃

V = (3.182 A) (37.7 Ω)

V = 120 V

230 V − 120 V = 110 V, the correct voltage for the lamps.

they are non-matter

explanation:

they are non-matter. basically, any type of energy or any abstract concept is an example of something that does not matter. other examples include dreams, magnetism, radio, and hate

scientific laws are the result of testing theories and proving them to be facts.