Eight holiday lights are connected in parallel as shown in Fig.
3. a. If the set is connected to a 120 V source, what is the current through each bulb if each bulb has an internal resistance of 1.8 k?
4. b. Determine the total resistance of the network.
5. c. Find the current drain from the supply.
6. d. What is the power delivered to each bulb?
7. e. Using the results of part (d), what is the power delivered by the source?
8. f. If one bulb burns out (that is, the filament opens up), what is the effect on the remaining bulbs? What is the effect on the source current? Why?
if they are in parallel the same 120 volts is on each
i = V/R = 120 volts /(1.8*10^3 Ohms) = 0.0667 amps through each bulb
total resistance = (1.8*10^3 )/8 = 225 Ohms
i = V/R = 120/225 = 0.534 amps. (0667*8 works too)
P = i V = 0.0667 * 120 = 8 watts per bulb
8*8 = 64 watts
There is no effect on the other bulbs. That is why they are wired in parallel these days.
There will be the same current through each so 7/8 of he original total current.
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To calculate the answers to these questions, we need to use Ohm's Law, which states that the current (I) flowing through a conductor is directly proportional to the voltage (V) applied to it and inversely proportional to the resistance (R) of the conductor.
a. The voltage across each bulb is the same when they are connected in parallel. Therefore, the voltage across each bulb is 120 V. The internal resistance of each bulb is given as 1.8 kΩ, which is the same as 1800 Ω. Using Ohm's Law (V = IR), we can find the current (I) flowing through each bulb as:
I = V / R = 120 V / 1800 Ω = 0.067 A
So, the current through each bulb is 0.067 A.
b. When multiple resistors are connected in parallel, the total resistance (R_total) can be found using the formula:
1 / R_total = 1 / R1 + 1 / R2 + 1 / R3 + ...
In this case, since each bulb has the same resistance (1800 Ω), we can find the total resistance by dividing the resistance of one bulb by the number of bulbs:
R_total = 1800 Ω / 8 = 225 Ω
c. The current drain from the supply can be calculated by dividing the total voltage by the total resistance:
I_supply = V / R_total = 120 V / 225 Ω = 0.533 A
So, the current drain from the supply is 0.533 A.
d. The power delivered to each bulb can be calculated using the formula:
P = V * I = 120 V * 0.067 A = 8.04 W
So, the power delivered to each bulb is 8.04 watts.
e. The power delivered by the source can be found by multiplying the total voltage by the total current:
P_source = V * I_supply = 120 V * 0.533 A = 63.96 W
So, the power delivered by the source is 63.96 watts.
f. If one bulb burns out (the filament opens up), it will increase the overall resistance of the network. As a result, the current flowing through the remaining bulbs will decrease. The effect on the source current is that it will also decrease because the total resistance has increased. This is because the bulbs are connected in parallel, so if one bulb fails, it will not affect the others directly. However, the remaining bulbs may receive a higher voltage than before, which can lead to a shorter lifespan or even failure of the remaining bulbs over time.
a. To find the current through each bulb, we can use Ohm's Law. Ohm's Law states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by the resistance (R) of the resistor.
In this case, the voltage across each bulb is 120 V (since they are connected in parallel), and the resistance of each bulb is 1.8 kΩ. So the equation becomes:
I = V / R = 120 V / 1.8 kΩ = 66.67 mA
Therefore, the current through each bulb is 66.67 mA.
b. The total resistance of a parallel network can be found by adding up the reciprocals of the individual resistances and then taking the reciprocal of the sum.
In this case, the resistance of each bulb is 1.8 kΩ. So the total resistance (R_total) can be calculated as:
1 / R_total = 1 / R1 + 1 / R2 + 1 / R3 + ... + 1 / Rn
where R1, R2, R3, ..., Rn are the resistances of each bulb.
R_total = 1 / (1 / R1 + 1 / R2 + 1 / R3 + ... + 1 / Rn)
For 8 bulbs in parallel, the equation becomes:
R_total = 1 / (1 / 1.8 kΩ + 1 / 1.8 kΩ + 1 / 1.8 kΩ + ... + 1 / 1.8 kΩ)
R_total = 1 / (8 / 1.8 kΩ) = 0.225 kΩ
Therefore, the total resistance of the network is 0.225 kΩ.
c. The current drain from the supply can be found by applying Ohm's Law again. Since the bulbs are connected in parallel, the total current drain (I_total) is equal to the sum of the currents through each bulb.
Since each bulb has a current of 66.67 mA (as calculated in part a), the total current drain is:
I_total = 8 * 66.67 mA = 533.36 mA
Therefore, the current drain from the supply is 533.36 mA.
d. The power delivered to each bulb can be calculated using the formula P = IV, where P is the power, I is the current, and V is the voltage across the bulb.
In this case, the voltage across each bulb is 120 V (given), and the current through each bulb is 66.67 mA (as calculated in part a). Therefore, the power delivered to each bulb is:
P = IV = (66.67 mA) * (120 V) = 8 W
Therefore, the power delivered to each bulb is 8 W.
e. Since the power delivered to each bulb is 8 W (as calculated in part d), the total power delivered by the source can be found by multiplying the power delivered to each bulb by the total number of bulbs.
Since there are 8 bulbs, the total power delivered by the source is:
Total power = Power per bulb * Number of bulbs = 8 W * 8 = 64 W
Therefore, the power delivered by the source is 64 W.
f. If one bulb burns out (the filament opens up), it will have no effect on the remaining bulbs. The remaining bulbs will continue to operate normally because they are still connected in parallel.
However, the effect on the source current will be different. When a bulb burns out, its resistance becomes infinite, effectively disconnecting it from the circuit. This will reduce the total resistance of the network (since there's one less bulb), which means the current drain from the supply will increase. The increased current will be distributed among the remaining bulbs, causing them to be brighter than before.
Therefore, the effect of one bulb burning out is that the remaining bulbs will become brighter, and the source current will increase.