### Loads - Scenario 3

SITUATION

Lamp 1 and Lamp 2, operating in parallel across 230 Volts, dissipate rated power at 100 Watts and 60 Watts respectively. How much power will each lamp dissipate if they are connected in series?

ANALYSIS

1.) Parallel operation

1.1.) Power:

P,1 = 100W

P,2 = 60W

1.2.) Resistance:

R,1 = (V)^2 / P,1

R,1 = (230V)^2 / 100W = 529 ohms

R,2 = (V)^2 / P,2

R,2 = (230V)^2 / 60W = 881.67 ohms

2.) Series operation

2.1.) Current:

V = i * (R,1 + R,2)

i = V / (R,1 + R,2)

i = 230V / [(529 + 881.67) ohms]

i = 0.16A

2.2.) Power:

P,1 = i^2 * R,1

P,1 = (0.16A)^2 * 529 ohms = 13.54W

P,2 = i^2 * R,2

P,2 = (0.16A)^2 * 881.67 ohms = 22.57W

CONCLUSION

When connected in series, Lamp 1 dissipates around 13.5 Watts while Lamp 2 dissipates around 22.6 Watts.

Both outputs decrease significantly, but Lamp 2 now has a higher power output than Lamp 1.

Lamp 1 and Lamp 2, operating in parallel across 230 Volts, dissipate rated power at 100 Watts and 60 Watts respectively. How much power will each lamp dissipate if they are connected in series?

ANALYSIS

1.) Parallel operation

1.1.) Power:

P,1 = 100W

P,2 = 60W

1.2.) Resistance:

R,1 = (V)^2 / P,1

R,1 = (230V)^2 / 100W = 529 ohms

R,2 = (V)^2 / P,2

R,2 = (230V)^2 / 60W = 881.67 ohms

2.) Series operation

2.1.) Current:

V = i * (R,1 + R,2)

i = V / (R,1 + R,2)

i = 230V / [(529 + 881.67) ohms]

i = 0.16A

2.2.) Power:

P,1 = i^2 * R,1

P,1 = (0.16A)^2 * 529 ohms = 13.54W

P,2 = i^2 * R,2

P,2 = (0.16A)^2 * 881.67 ohms = 22.57W

CONCLUSION

When connected in series, Lamp 1 dissipates around 13.5 Watts while Lamp 2 dissipates around 22.6 Watts.

Both outputs decrease significantly, but Lamp 2 now has a higher power output than Lamp 1.

## Comments

## Post a Comment