If a vacuum tube with a filament rating of 0.25 A and 5 V is operated from a 6 volt battery, what value of resistor is necessary?

If a vacuum tube with a filament rating of 0.25 A and 5 V is operated from a 6 volt battery, what value of resistor is necessary?

• A. 4 ohms
• B. 5 ohms
• C. 10 ohms
• D. 2 ohms

Answer: (A)

To determine the value of the resistor needed when operating a vacuum tube with a filament rating of 0.25 A and 5 V from a 6 V battery, it's important to first understand the voltage difference that the resistor must accommodate. The filament requires 5 V, and the supply voltage is 6 V, creating an excess voltage that must be dropped across the resistor. The voltage drop across the resistor can be calculated as: \[ V_{resistor} = V_{supply} - V_{filament} \] \[ V_{resistor} = 6 V - 5 V = 1 V \] Next, using Ohm's Law (V = IR), where V is the voltage across the resistor, I is the current through the circuit, and R is the resistance, we can rearrange the formula to solve for R: \[ R = \frac{V_{resistor}}{I} \] Substituting the known values: \[ R = \frac{1 V}{0.25 A} = 4 \, \text{ohms} \] This calculation shows that to drop the necessary 1 volt across the resistor while maintaining a current of 0.25 A, a resistor

To determine the value of the resistor needed when operating a vacuum tube with a filament rating of 0.25 A and 5 V from a 6 V battery, it's important to first understand the voltage difference that the resistor must accommodate.

The filament requires 5 V, and the supply voltage is 6 V, creating an excess voltage that must be dropped across the resistor. The voltage drop across the resistor can be calculated as:

[ V_{resistor} = V_{supply} - V_{filament} ]

[ V_{resistor} = 6 V - 5 V = 1 V ]

Next, using Ohm's Law (V = IR), where V is the voltage across the resistor, I is the current through the circuit, and R is the resistance, we can rearrange the formula to solve for R:

[ R = \frac{V_{resistor}}{I} ]

Substituting the known values:

[ R = \frac{1 V}{0.25 A} = 4 , \text{ohms} ]

This calculation shows that to drop the necessary 1 volt across the resistor while maintaining a current of 0.25 A, a resistor