I measure a voltage drop of 30mv over a 2.5 ohm resistor. How long will a 3Ah battery last? Write your ansser 'XXX hours'
I measure a voltage drop of 30mv over a 2.5 ohm resistor. How long will a 3Ah battery last? Write your ansser 'XXX hours'
First, let's determine the current flowing through the resistor using Ohm's Law, which states: \[ V = I \times R \] Given: - Voltage drop (V) = 30 mV = 0.03 V - Resistance (R) = 2.5 ohms Rearranging the formula to solve for current (I): \[ I = \frac{V}{R} \] Substituting the given values: \[ I = \frac{0.03 \text{ V}}{2.5 \text{ ohms}} = 0.012 \text{ A} = 12 \text{ mA} \] Next, we need to determine how long a 3 Ah battery will last when supplying this current. The battery capacity in ampere-hours (Ah) provides a measure of how much current the battery can supply over a given period. Given: - Battery capacity = 3 Ah - Current (I) = 0.012 A (12 mA) The time (T) in hours can be calculated using the formula: \[ T = \frac{\text{Battery Capacity}}{\text{Current}} \] Substituting the given values: \[ T = \frac{3 \text{ Ah}}{0.012 \text{ A}} = 250 \text{ hours} \] Therefore, the battery will last: \[ 250 \text{ hours} \]
Testing if the previous output matches the regex 250.{,10}hours: True
I measure a voltage drop of 30mv over a 2.5 ohm resistor. How long will a 3Ah battery last? Write your ansser 'XXX hours'
I measure a voltage drop of 30mv over a 2.5 ohm resistor. How long will a 3Ah battery last? Write your ansser 'XXX hours'
To determine how long the 3Ah battery will last, we need to calculate the current flowing through the resistor and then use that to find the battery life. 1. **Calculate the current using Ohm's law:** \[ V = IR \] Given: - Voltage drop \( V = 30 \text{ mV} = 0.03 \text{ V} \) - Resistance \( R = 2.5 \text{ ohms} \) Rearrange Ohm's law to solve for the current \( I \): \[ I = \frac{V}{R} = \frac{0.03 \text{ V}}{2.5 \text{ ohms}} = 0.012 \text{ A} = 12 \text{ mA} \] 2. **Calculate the battery life:** The battery capacity is given as 3Ah, which means it can supply 3 amperes for 1 hour. To find out how long it can supply 12 mA, we use the formula: \[ \text{Battery life (hours)} = \frac{\text{Battery capacity (Ah)}}{\text{Current (A)}} \] Given: - Battery capacity \( = 3 \text{ Ah} \) - Current \( = 0.012 \text{ A} \) Substitute the values: \[ \text{Battery life (hours)} = \frac{3 \text{ Ah}}{0.012 \text{ A}} = 250 \text{ hours} \] So, the battery will last **250 hours**.
Testing if the previous output matches the regex 250.{,10}hours: True
I measure a voltage drop of 30mv over a 2.5 ohm resistor. How long will a 3Ah battery last? Write your ansser 'XXX hours'
I measure a voltage drop of 30mv over a 2.5 ohm resistor. How long will a 3Ah battery last? Write your ansser 'XXX hours'
To determine how long a 3Ah battery will last given a voltage drop of 30mV over a 2.5 ohm resistor, follow these steps: 1. **Calculate the current (I):** Using Ohm's Law, \( V = I \times R \), where \( V \) is the voltage drop, \( I \) is the current, and \( R \) is the resistance. Given: \[ V = 30 \text{ mV} = 0.030 \text{ V} \] \[ R = 2.5 \text{ ohms} \] Rearrange the formula to solve for \( I \): \[ I = \frac{V}{R} = \frac{0.030 \text{ V}}{2.5 \text{ ohms}} = 0.012 \text{ A} = 12 \text{ mA} \] 2. **Calculate the battery life:** The battery capacity is given as 3Ah (Ampere-hours). To find out how long the battery will last, use the formula: \[ \text{Battery Life (hours)} = \frac{\text{Battery Capacity (Ah)}}{\text{Current (A)}} \] Given: \[ \text{Battery Capacity} = 3 \text{ Ah} \] \[ \text{Current} = 0.012 \text{ A} \] Calculate the battery life: \[ \text{Battery Life} = \frac{3 \text{ Ah}}{0.012 \text{ A}} = 250 \text{ hours} \] Therefore, the battery will last **250 hours**.
Testing if the previous output matches the regex 250.{,10}hours: True
I measure a voltage drop of 30mv over a 2.5 ohm resistor. How long will a 3Ah battery last? Write your ansser 'XXX hours'
I measure a voltage drop of 30mv over a 2.5 ohm resistor. How long will a 3Ah battery last? Write your ansser 'XXX hours'
First, let's determine the current flowing through the resistor using Ohm's Law, which states \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. Given: - Voltage drop \( V = 30 \text{ mV} = 0.030 \text{ V} \) - Resistance \( R = 2.5 \text{ ohms} \) The current \( I \) can be calculated as: \[ I = \frac{V}{R} = \frac{0.030 \text{ V}}{2.5 \text{ ohms}} = 0.012 \text{ A} \] Now that we have the current, we can determine how long the battery will last. The battery capacity is given in ampere-hours (Ah), which means it can supply a certain current for a certain number of hours. Given: - Battery capacity = 3 Ah - Current = 0.012 A The duration \( T \) the battery will last can be calculated as: \[ T = \frac{\text{Battery capacity}}{\text{Current}} = \frac{3 \text{ Ah}}{0.012 \text{ A}} = 250 \text{ hours} \] So, the battery will last \( 250 \text{ hours} \).
Testing if the previous output matches the regex 250.{,10}hours: True
I measure a voltage drop of 30mv over a 2.5 ohm resistor. How long will a 3Ah battery last? Write your ansser 'XXX hours'
I measure a voltage drop of 30mv over a 2.5 ohm resistor. How long will a 3Ah battery last? Write your ansser 'XXX hours'
To determine how long a 3Ah battery will last, we need to calculate the current being drawn and then use that to find the time. 1. **Calculate the current (I) using Ohm's Law:** \[ I = \frac{V}{R} \] where \( V \) is the voltage drop and \( R \) is the resistance. Given: \[ V = 30 \text{ mV} = 0.030 \text{ V} \] \[ R = 2.5 \text{ ohms} \] Substituting the values: \[ I = \frac{0.030 \text{ V}}{2.5 \text{ ohms}} = 0.012 \text{ A} = 12 \text{ mA} \] 2. **Calculate the battery life:** The battery life can be calculated using the formula: \[ \text{Battery life (hours)} = \frac{\text{Battery capacity (Ah)}}{\text{Current (A)}} \] Given: \[ \text{Battery capacity} = 3 \text{ Ah} \] \[ \text{Current} = 0.012 \text{ A} \] Substituting the values: \[ \text{Battery life} = \frac{3 \text{ Ah}}{0.012 \text{ A}} = 250 \text{ hours} \] So, the battery will last \( 250 \) hours.
Testing if the previous output matches the regex 250.{,10}hours: True