You are in the lab with a 10 V battery. You have a 10 W light bulb and you want to put 1/2 A of current through the light bulb and 5 V across the light bulb. This will allow the light bulb to have a brightness proportional to 2.5 W. Assuming any resistors that you need, design and build a circuit that delivers the required current and voltage to the bulb.

To design a circuit that delivers the required current and voltage to the light bulb, we need to calculate the resistance (R) required for the circuit.

Using Ohm's Law (V = IR), we can rearrange the formula to solve for resistance:
R = V / I

Given that the voltage across the bulb should be 5 V and the desired current is 1/2 A, we can substitute these values into the equation:
R = 5 V / 0.5 A

R = 10 Ω

Since we need to use a 10 W bulb, which is designed to operate at a certain power, we can use the formula for power (P) to calculate the resistance required to achieve the desired brightness:
P = V^2 / R

Substituting the known values:
10 W = (5 V)^2 / R

Solving for R, we get:
R = (5 V)^2 / 10 W

R = 25 Ω

Therefore, the circuit should have a resistance of 25 Ω in series with a 10 W light bulb to deliver the required current and voltage while maintaining the desired brightness.

To build the circuit, you can follow these steps:

1. Connect the positive terminal of the battery to one end of the light bulb.
2. Connect the other end of the bulb to one end of a 25 Ω resistor.
3. Connect the other end of the resistor to the negative terminal of the battery.

This configuration will provide the desired current of 1/2 A through the light bulb and the desired voltage of 5 V across the bulb, allowing it to operate at the desired brightness of 2.5 W.