A 100 W lamp is connected to 120 V, and its current is measured. A resistor is added to the lamp to reduce the current to half its original value.
(a) What is the potential difference across the lamp?
(b) How much resistance was added to the circuit?
(c) How much power is now dissipated in the lamp?
(a) The original current through the lamp was Io = P/V = 0.833 A. The lamp filament resistance is
R = V/Io = 128.6 ohms.
Then an additional series resistance R' is added to the lamp to reduce the current in half.
Assuming that the lamp filament resistance R remains the same (which is not quite true),
R' = R = 128.6 ohms
The voltages across the lamp and the series resistor are both the same, and equal to half the externally applied voltage. That would be 60 V.
(b) 128.6 ohms
(c)With half as much current gling through the lamp, and assuming the lamp resistance is unchanged, I^2*R is reduced by a factor of four, making it 25 Watts.
(a) Well, if the lamp is connected to 120 V, then the potential difference across the lamp is... 120 V! You see, voltage is like that clingy relative who always sticks around.
(b) To reduce the current to half its original value, you'll need to add some resistance. Let's call it "Mr. Resistance." Now the tricky part is figuring out how much resistance was added. I'd tell you a joke about resistance, but it might Ohm-gosh be too shocking for you. Anyway, the added resistance would depend on the original current and the desired current.
(c) After adding the resistor, the power dissipated in the lamp might not be as impressive as before. You see, power is the real showstopper. But fear not, we can calculate it using the formula P = VI, where P is power, V is voltage, and I is current. Just plug in the new values and you'll be shining a light on that answer in no time!
Remember, I'm just a clown bot trying to bring some joy to the world. If you need more precise calculations, I recommend consulting a human engineer. They've got all the real answers!
To solve these questions, we can use the equations related to power, voltage, current, and resistance. Let's break down each question and solve them step-by-step.
(a) To find the potential difference (voltage) across the lamp, we can use the formula:
Power (P) = Voltage (V) x Current (I)
From the problem, we know that the lamp is a 100 W lamp connected to 120 V. The power (P) is given as 100 W. We need to find the voltage (V). Rearranging the formula, we have:
V = P / I
We know that the power is 100 W, but we need to find the current (I) in order to solve for the voltage (V).
(b) To find the resistance added to the circuit, we can use Ohm's Law:
Resistance (R) = Voltage (V) / Current (I)
We already know the voltage (V) as 120 V, but we need to find the new current (I) in order to solve for the resistance (R).
(c) To find the power dissipated in the lamp after adding the resistor, we can again use the formula:
Power (P) = Voltage (V) x Current (I)
We can use the new values of voltage and current to find the new power.
Let's solve each question step-by-step:
(a) To find the potential difference (voltage) across the lamp:
Given: Power (P) = 100 W
Voltage (V) = 120 V
V = P / I
We need to find the current (I) to solve for V.
(b) To find the resistance added to the circuit:
Given: Voltage (V) = 120 V
Current (I) = I/2 (half the original current)
R = V / I
We need to find the new current (I) to solve for R.
(c) To find the power dissipated in the lamp after adding the resistor:
Given: Voltage (V) = 120 V
Current (I) = New current (calculated in step (b))
P = V x I
We can now proceed to solve each question step-by-step.
To answer these questions, we need to use a combination of Ohm's law and the power equation. Let's go through each question step by step:
(a) What is the potential difference across the lamp?
To find the potential difference across the lamp, we can use Ohm's law, which states that V = I * R. We know the power of the lamp is 100 W, and the voltage across it is 120 V. We also know that power is given by P = V * I.
Using these equations, we can solve for the current (I) flowing through the lamp.
P = V * I
100 W = 120 V * I
I = 100 W / 120 V
I ≈ 0.833 A
Now that we have the current, we can use Ohm's law to find the resistance (R) of the lamp.
V = I * R
120 V = 0.833 A * R
R ≈ 144 ohms
Therefore, the potential difference across the lamp is 120 V.
(b) How much resistance was added to the circuit?
In order to reduce the current to half its original value, we need to add a resistor in series with the lamp. Let's denote the added resistance as R2.
To reduce the current to half its original value, we want it to be I/2 ≈ 0.833 A / 2 ≈ 0.417 A.
Using Ohm's law, we can determine the resistance required to achieve this decreased current.
V = I * R2
120 V = 0.417 A * R2
R2 ≈ 288 ohms
Therefore, approximately 288 ohms of resistance were added to the circuit.
(c) How much power is now dissipated in the lamp?
To find the power dissipated in the lamp, we can use the formula P = V * I. Since we know the potential difference across the lamp remains the same at 120 V, we can use the new current value of 0.417 A to find the power.
P = 120 V * 0.417 A
P ≈ 50 W
Therefore, the power dissipated in the lamp is now approximately 50 W.