Two 200 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. How much power is dissipated by each bulb?
in normal operation of a bulb
200 Watts = V^2/R = 120^2/R
so R = 72 Ohms
now we have two in series so V = 120/2 = 60 volts
again watts power = V^2/R = 60^2/72
= 50 watts
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To determine the power dissipated by each lightbulb, we need to understand the concept of series circuit and apply the appropriate formulas.
In a series circuit, the total resistance (R_total) is equal to the sum of the individual resistances. Similarly, the total voltage (V_total) is equal to the sum of the individual voltages.
First, let's calculate the total resistance (R_total) in the circuit. Since the lightbulbs are connected in series, the resistance of each bulb adds up:
R_total = R1 + R2
Next, let's calculate the total voltage (V_total) in the circuit, which is equal to the supply voltage:
V_total = 120 V
Now we can use Ohm's Law to calculate the current (I) flowing through the circuit:
I = V_total / R_total
Finally, we can calculate the power dissipated by each lightbulb using the formula:
P = I^2 * R
Since the resistance (R) of the lightbulbs is the same, we can calculate the power dissipated by each bulb as:
P_each bulb = (I^2 * R_total) / 2
Let's go step by step:
1. Calculate the resistance (R_total) in the circuit:
- Since the lightbulbs are in series, the resistance adds up:
R_total = R1 + R2 = 200 W + 200 W = 400 W
2. Calculate the current (I) flowing through the circuit:
- Using Ohm's Law: I = V_total / R_total
I = 120 V / 400 W = 0.3 A
3. Calculate the power dissipated by each lightbulb:
- Using the formula: P_each bulb = (I^2 * R_total) / 2
P_each bulb = (0.3 A)^2 * 400 W / 2 = 0.09 * 400 W / 2 = 18 W
Therefore, each lightbulb dissipates 18 W of power.