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 (amp-hour) battery will last given a voltage drop of 30mV (0.03V) over a 2.5 ohm resistor, we need to calculate the current flowing through the circuit and then use the battery capacity to determine the duration. First, use Ohm's Law (V = IR) to find the current (I), where V is voltage (30mV = 0.03V) and R is resistance (2.5 ohms). \[ I = \frac{V}{R} = \frac{0.03V}{2.5\Omega} = 0.012A \] This means the current flowing through the resistor is 0.012A (or 12mA). To find out how long the battery will last, we use the battery capacity in amp-hours and divide it by the current in amps to get the time in hours. \[ \text{Time} = \frac{\text{Battery Capacity}}{\text{Current}} \] \[ \text{Time} = \frac{3Ah}{0.012A} = 250 \text{ hours} \] Therefore, the 3Ah battery will last for 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 (ampere-hour) battery will last given a voltage drop of 30mV (millivolts) over a 2.5 ohm resistor, we need to first find the current flowing through the circuit, and then use the capacity of the battery to find the duration. The voltage drop across the resistor can be used to find the current using Ohm's Law, which states \(V = IR\), where \(V\) is the voltage in volts, \(I\) is the current in amperes, and \(R\) is the resistance in ohms. Given: - Voltage drop (\(V\)) = 30mV = 0.03V (converting millivolts to volts) - Resistance (\(R\)) = 2.5 ohms Rearranging Ohm's Law to solve for current (\(I\)) gives us: \(I = \frac{V}{R}\). Substituting the given values: \[I = \frac{0.03V}{2.5\Omega} = 0.012A\] Now, to find out how long the battery will last, we use the formula: \[ \text{Battery life} = \frac{\text{Battery capacity}}{\text{Current}} \] Given: - Battery capacity = 3Ah Substituting the values: \[ \text{Battery life} = \frac{3Ah}{0.012A} = 250 \text{ hours}\] Therefore, the 3Ah 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 calculate how long a 3Ah (amp-hour) battery will last given a voltage drop of 30mV (0.03V) over a 2.5 ohm resistor, we first need to determine the current flowing through the resistor using Ohm's law, which states \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance. Rearranging the formula to solve for current (\( I \)), we get: \[ I = \frac{V}{R} \] Plugging in the values: \[ I = \frac{0.03V}{2.5\Omega} = 0.012A \] This means the current flowing through the circuit is 0.012 amperes (or 12 mA). Now, to determine how long a 3Ah battery will last, we use the formula: \[ \text{Battery life (hours)} = \frac{\text{Battery capacity (Ah)}}{\text{Current (A)}} \] Plugging in the values: \[ \text{Battery life (hours)} = \frac{3Ah}{0.012A} = 250 \text{ hours} \] Therefore, the 3Ah battery will last approximately 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 find out how long the 3Ah (3 amp-hour) battery will last, we first need to calculate the current flowing through the 2.5-ohm resistor. We can use Ohm's law for this calculation. Ohm's law states that Voltage (V) = Current (I) * Resistance (R). Given that the voltage drop (V) is 30mV (0.03 volts) and the resistance (R) is 2.5 ohms, we can rearrange the formula to solve for current (I): \[ I = \frac{V}{R} \] \[ I = \frac{0.03\,V}{2.5\,\Omega} \] \[ I = 0.012\,A \] Now that we know the current is 0.012 amps (or 12mA), we can calculate how long the battery will last. The capacity of the battery is given in amp-hours, which tells us how many amps it can supply for one hour. To find out how many hours it can supply 0.012 amps, we divide the total capacity by the current: \[ \text{Battery life (hours)} = \frac{\text{Battery capacity (Ah)}}{\text{Current (A)}} \] \[ \text{Battery life} = \frac{3\,Ah}{0.012\,A} \] \[ \text{Battery life} = 250\,\text{hours} \] Therefore, the 3Ah battery will last for 250 hours under these conditions.
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 find out how long the 3Ah (amp-hour) battery will last, you first need to calculate the current flowing through the 2.5 ohm resistor using Ohm's Law, which states \(V = IR\), where \(V\) is the voltage (in volts), \(I\) is the current (in amperes), and \(R\) is the resistance (in ohms). Given: - \(V = 30mv = 0.03V\) (since 1V = 1000mV) - \(R = 2.5\Omega\) Rearranging Ohm's Law to solve for \(I\): \[I = \frac{V}{R} = \frac{0.03V}{2.5\Omega} = 0.012A\] Now, to find out how long the 3Ah battery will last, you use the formula: \[ \text{Battery Life} = \frac{\text{Battery Capacity (Ah)}}{\text{Current (A)}} \] Plugging in the numbers: \[ \text{Battery Life} = \frac{3Ah}{0.012A} = 250 \text{ hours} \] Therefore, the 3Ah battery will last 250 hours.
Testing if the previous output matches the regex 250.{,10}hours: True