Question

[Knight, 28.50] A lightbulb is in series with a 2.0Ω resistor. The lightbulb dissipates 10 W when this series circuit is connected to a 9.0 V battery. What is the current through the lightbulb? There are two possible answers; give the smaller of them on D2L.

Answer 1

To determine the current through the lightbulb in series with a 2.0Ω resistor and a battery providing 9.0 V, we calculate the total resistance including the bulb and then apply the power formula. The lightbulb's resistance is found to be 6.1Ω leading to a current approximately equal to 1.28 A.

Step-by-step explanation:

To find the current through the lightbulb, we can apply the concept of power and Ohm's law. Power (P) is the rate at which energy is transferred and can be calculated for a resistor in a circuit using the formula P = I2R, where I is the current and R is the resistance.

Given that the lightbulb dissipates 10 W of power and is in series with a 2.0Ω resistor, and the total voltage provided by the battery is 9.0 V, we can set up an equation to find the current flowing through the lightbulb. First, let's calculate the total resistance in the circuit as the sum of the lightbulb's resistance (Rbulb) and the 2Ω resistor.

The total power dissipated in the series circuit is also the power dissipated by the lightbulb (since the components are in series, they have the same current). By using the power formula P = V2 / Rtotal, we can solve for the lightbulb's resistance:

Rtotal = V2 / P = (9.0V)2 / 10W = 8.1Ω

Since we know the resistor in series is 2Ω, the lightbulb's resistance is:

Rbulb = Rtotal - 2.0Ω = 8.1Ω - 2.0Ω = 6.1Ω

Now, we can calculate the current using the power formula, re-arranged to I = √(P / R):

I = √(10W / 6.1Ω) ≈ 1.28 A

Therefore, the current through the lightbulb in this series circuit is approximately 1.28 A.

Answer 2

The current through the lightbulb in the series circuit is approximately 0.2951 A.

To find the current through the lightbulb in the series circuit, we can use Ohm's law and the power equation.

First, let's calculate the total resistance in the circuit. Since the lightbulb and the 2.0Ω resistor are in series, we can simply add their resistances together:

Total resistance = resistance of lightbulb + resistance of 2.0Ω resistor
Total resistance = 28.50Ω + 2.0Ω
Total resistance = 30.50Ω

Next, we can use Ohm's law to find the current through the circuit:

Current (I) = Voltage (V) / Resistance (R)

Given that the voltage is 9.0V and the resistance is 30.50Ω, we can substitute these values into the equation:

Current (I) = 9.0V / 30.50Ω
Current (I) ≈ 0.2951 A