010 (part 1 of 5) 10.0 points
A string of 26 identical Christmas tree lights
are connected in series to a 120 V source. The
string dissipates 63 W.
What is the equivalent resistance of the
light string?
Answer in units of Ω.
011 (part 2 of 5) 10.0 points
What is the resistance of a single light?
Answer in units of Ω.
012 (part 3 of 5) 10.0 points
What power is dissipated in a single light?
Answer in units of W.
013 (part 4 of 5) 10.0 points
One of the bulbs burns out. The lamp has a
wire that shorts out the bulb filament when it
burns out, dropping the resistance of the bulb
to zero.
What is the resistance Rnew of the light
string now?
Answer in units of Ω.
014 (part 5 of 5) 10.0 points
Find the power Pnew dissipated by the string
now.
Answer in units of W.
I will be happy to critque your thinking.
To solve this problem, we need to use the formulas related to electrical circuits.
010:
The equivalent resistance of a series circuit can be found by summing the resistances of each component. In this case, all the lights are connected in series, so the equivalent resistance of the light string is equal to the sum of the resistances of the individual lights.
011:
If the equivalent resistance of the light string is known, we can find the resistance of a single light by dividing the equivalent resistance by the number of lights in the string.
012:
The power dissipated in a component can be calculated using the formula: power = (current)^2 * resistance. In this case, we need to find the power dissipated in a single light.
013:
When one of the bulbs burns out and the resistance drops to zero, the remaining bulbs in the string will still have resistance, but the equivalent resistance of the string will change. We need to calculate the new resistance of the light string with one bulb burned out.
014:
Now that we know the new resistance of the light string, we can find the power dissipated by the string by using the same formula as in 012.
Now, let's solve each part of the problem step by step.
010:
The formula for equivalent resistance in a series circuit is:
R_eq = R_1 + R_2 + ... + R_n
In this case, all the lights are identical, so we can represent the equivalent resistance of the string as:
R_eq = R + R + ... + R (26 times)
To calculate the equivalent resistance, we need to know the value of one resistance since they are all the same. We can use the power dissipated by the string and the given voltage to find the resistance.
Given: power = 63 W and voltage = 120 V
The formula for power in terms of voltage and resistance is:
power = (voltage)^2 / resistance
Rearranging the formula, we have:
resistance = (voltage)^2 / power
Substituting the given values, we get:
resistance = (120 V)^2 / 63 W
Calculating this expression gives us the value of one resistance.
011:
To find the resistance of a single light, we divide the equivalent resistance (from part 010) by the number of lights in the string, which is 26 in this case.
Resistance of a single light = R_eq / 26
012:
To find the power dissipated in a single light, we need to use the formula for power again:
power = (current)^2 * resistance
However, we don't know the current flowing through the circuit. But we can find it using Ohm's Law:
current = voltage / R_eq
Substituting this expression for current into the power formula, we get:
power = (voltage^2 / R_eq^2) * resistance
Now, substitute the known values to calculate the power dissipated in a single light.
013:
When one of the bulbs burns out, its resistance drops to zero. This will change the equivalent resistance of the light string.
To calculate the new resistance, we can:
1. Remove the burnt-out bulb from the circuit.
2. Calculate the new equivalent resistance as we did in part 010, but with 25 lights instead of 26. This gives us the resistance when the burnt-out bulb is removed.
Rnew = R + R + ... + R (25 times)
014:
Now that we know the new resistance of the light string, we can calculate the power dissipated by the string using the formula:
power = (voltage^2 / R_new^2) * resistance
Substituting the known values, we can calculate the new power dissipated by the string.
By following these steps, you can find the answers to each part of the problem.