Question

On what factors does the resistance of a conductor depend? An electric bulb operated at 80 volts takes 10 amperes of current. How much resistance should be connected in series with the bulb to use it at 240 volts so that it takes the same current?

Hint:

When using electrical devices at higher voltages, resistances in series can help regulate the current to the desired value by limiting the total current according to Ohm's law: \( I = \frac{V}{R} \).

Answer 1

Step 1: Factors affecting the resistance of a conductor.
The resistance of a conductor depends on the following factors:
1.
Length of the conductor (L): The resistance is directly proportional to the length of the conductor. As the length increases, resistance increases. \[ R \propto L \]
2.
Cross-sectional area (A): The resistance is inversely proportional to the cross-sectional area of the conductor. As the area increases, the resistance decreases. \[ R \propto \frac{1}{A} \]
3.
Material of the conductor: The resistance depends on the material's resistivity, denoted by \( \rho \). Each material has a characteristic resistivity that determines its resistance. \[ R = \rho \frac{L}{A} \] 4.
Temperature of the conductor: The resistance of most conductors increases with temperature.
Step 2: Calculating the required resistance to use the bulb at 240 volts. We are given:
- The voltage \( V_1 = 80 \, \text{V} \),
- The current \( I_1 = 10 \, \text{A} \),
- The voltage \( V_2 = 240 \, \text{V} \),
- The current \( I_2 = I_1 = 10 \, \text{A} \) (since we want the same current).
First, calculate the resistance of the bulb at 80 V using Ohm's law: \[ R_1 = \frac{V_1}{I_1} = \frac{80}{10} = 8 \, \Omega \] Now, when the bulb is used at 240 V, we want the same current (10 A) to flow through the bulb. The total resistance in the circuit must be such that: \[ R_{\text{total}} = \frac{V_2}{I_2} = \frac{240}{10} = 24 \, \Omega \] Since the resistance of the bulb \( R_1 \) is already 8 \( \Omega \), the resistance \( R_{\text{series}} \) to be connected in series must be: \[ R_{\text{series}} = R_{\text{total}} - R_1 = 24 - 8 = 16 \, \Omega \]
Step 3: Conclusion.
The resistance that should be connected in series with the bulb to use it at 240 V and maintain the same current of 10 A is
16 \( \Omega \).