(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.

no

sodium is solid at rt