#1. Ten Christmas lights are connected in series and plugged into a 120 V wall outlet. Each light is identical and can be thought of as a resistor.
a) What should the value of this resistance be if the total string is to dissipate 50 W?
b) How much power is dissipated in each resistor?
c) How much current flows in each resistor?
#2. Ten Christmas lights are now connected in parallel. Repeat problem #1 for this new design.
#3. Which string lights (#1 or #2) makes a better consumer product? Why?
I will be happy to critique your thinking.
#1. Ten Christmas lights connected in series:
a) To find the value of resistance required for the total string to dissipate 50 W, we can use the formula P = (V^2) / R, where P is power, V is voltage, and R is resistance.
Given that the power is 50 W and the voltage is 120 V, we can rearrange the formula to find the resistance:
R = (V^2) / P = (120^2) / 50 = 288 Ω
Therefore, the resistance needed for the total string to dissipate 50 W is 288 Ω.
b) In a series circuit, the current flowing through each resistor (light) is the same. Therefore, the power dissipated in each resistor can be calculated by using the formula P = (I^2) * R, where I is the current and R is the resistance.
Since the total power is 50 W and there are ten resistors, the power dissipated in each resistor is:
P = 50 W / 10 = 5 W
Therefore, each resistor (light) should dissipate 5 W of power.
c) In a series circuit, the current is the same throughout the circuit. Therefore, the current flowing in each resistor is the total current divided by the number of resistors.
To find the current flowing in each resistor, we can use Ohm's Law: I = V / R, where I is the current, V is the voltage (120 V), and R is the resistance (288 Ω).
I = 120 V / 288 Ω ≈ 0.416 A
Therefore, approximately 0.416 A of current flows through each resistor.
#2. Ten Christmas lights connected in parallel:
If the ten Christmas lights are connected in parallel, the analysis will be different from the series case.
a) In a parallel circuit, the total resistance is given by the reciprocal of the sum of the reciprocals of the individual resistances. Therefore, the total resistance in a parallel circuit is less than the smallest individual resistance.
Since the lights are identical, the resistance of each is 288 Ω as determined in problem #1. Therefore, the total resistance is:
1/R_total = 1/R1 + 1/R2 + ... + 1/R10 = 1/288 + 1/288 + ... + 1/288 = 10/288
R_total = 288 / 10 = 28.8 Ω
Therefore, the total resistance in the parallel circuit is 28.8 Ω.
b) In a parallel circuit, the voltage across each resistor is the same. Therefore, the power dissipated in each resistor can be calculated using the formula P = (V^2) / R, where V is the voltage and R is the resistance.
Since the total power is 50 W and there are ten resistors, the power dissipated in each resistor is:
P = 50 W / 10 = 5 W
Therefore, each resistor (light) should dissipate 5 W of power, same as in the series case.
c) In a parallel circuit, the current flowing through each resistor can be calculated using Ohm's Law: I = V / R, where I is the current, V is the voltage (120 V), and R is the resistance (28.8 Ω).
I = 120 V / 28.8 Ω ≈ 4.167 A
Therefore, approximately 4.167 A of current flows through each resistor.
#3. Which string lights (#1 or #2) makes a better consumer product? Why?
The parallel circuit (#2) makes a better consumer product compared to the series circuit (#1).
In a parallel circuit, if one light fails or is removed, the other lights will continue to work. This is because each light has its own separate path for current flow, and removing one does not interrupt the flow to the others. On the other hand, in a series circuit, if one light fails or is removed, the circuit will be broken, and all the lights will go out.
Furthermore, in a parallel circuit, the voltage across each light is the same, meaning that all the lights would shine with the same intensity. In a series circuit, the voltage is divided among the lights, resulting in reduced brightness for each light.
Overall, a parallel circuit provides greater reliability and consistent brightness compared to a series circuit, making it a better consumer product for Christmas lights.