Question

How can you light a \(1.0-\mathrm{W}, 1.5-\mathrm{V}\) bulb with your \(12.0-V\) car battery?

Short answer

Answer: To safely connect the bulb to the car battery, a 15.67-ohm resistor with a power rating of at least 7.04 Watts is required.

Step-by-step solution

Calculate the current needed for the bulb

To calculate the current needed for the 1.0-W, 1.5-V bulb, we can use the formula: Power = Voltage * Current Rearranging this equation for the current, we get: Current = Power / Voltage Now, we can plug in the power and voltage values of the bulb: Current = \(\frac{1.0 \text{ W}}{1.5 \text{ V}} = 0.67 \text{ A}\) The bulb requires 0.67 A to operate.

Determine the appropriate resistor value

Since we have a 12.0-V car battery and a 1.5-V bulb, we need to drop the remaining 10.5 V across the resistor. Using Ohm's law, we can find the resistor value: Voltage = Current * Resistance Rearranging this equation for the resistance, we get: Resistance = Voltage / Current Plugging in the values, we get: Resistance = \(\frac{10.5 \text{ V}}{0.67 \text{ A}} = 15.67 \text{ }\Omega\) We will need a 15.67-ohm resistor to reduce the voltage from 12.0 V to 1.5 V.

Calculate the power dissipation of the resistor

To find out if the resistor can handle the power dissipation, we can use the formula: Power = Voltage * Current Plugging in the values for the resistor, we get: Power = \(10.5 \text{ V} * 0.67 \text{ A} = 7.04 \text{ W}\) The resistor will need to dissipate 7.04 W.

Connect the resistor and the bulb

Now, we can connect the 15.67-ohm resistor in series with the1.0-W, 1.5-V bulb to the car battery. Make sure to use a suitable resistor that can handle the power dissipation of 7.04 W. This will ensure that the bulb receives the correct voltage and current to operate without damaging it.

Key concepts

Ohm's Law Application

When tackling electrical circuit problems, understanding Ohm's Law is crucial. It defines the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit, neatly wrapped up in the formula:
\( V = I \times R \).
By rearranging this fundamental formula, you can solve for any one of these variables provided you know the other two. For instance, to solve for current, you would use \( I = \frac{V}{R} \).
In the example of powering a 1.5V bulb with a 12.0V car battery, Ohm's Law aids in determining the right resistor that can drop the excess voltage down to what the bulb requires. Calculating the appropriate resistance to use ensures the bulb receives just enough power, which is a skillful application of Ohm's Law in designing practical circuits.

Power Dissipation Calculation

Power dissipation represents how much energy per unit time is converted to heat in circuit components. It's not only a factor in device performance but also in their safety and longevity. The formula for power dissipation is often given by
\( P = V \times I \),
where P is power in watts, V is voltage in volts, and I is current in amperes.
In the exercise where a resistor's power dissipation is calculated, understanding the significance of this equation helps to ensure that the resistor can handle the energy without overheating. If a resistor dissipates more power than it can safely manage, it could potentially fail or cause damage to the circuit. Thus, calculating the power dissipation is a critical step in circuit design.

Series Circuit Design

Creating a series circuit involves several components connected end-to-end so the same current flows through each component. In the case of the 1.5V bulb and 12.0V car battery, they are connected in series with a resistor to regulate the voltage.
Designing a series circuit requires an understanding of how voltages add up and how to use resistors to create specific voltage drops. The total voltage in the circuit is the sum of all individual voltage drops, so the chosen resistors need to be calculated precisely to maintain integrity and function of the components.

• Use Ohm's Law to calculate individual resistor values.
• Remember that the same current flows through every part of a series circuit.
• Check each component's power rating to prevent overheating.

Designing a series circuit such as the one in the exercise teaches students the practical application of the concepts of resistance and potential difference in real-world situations.