A 8 V flashlight battery has an internal resistance of 1.7 Ω when new, and 8 Ω when old. Imagine this is a resistor in series with the battery mounted INSIDE the battery itself, which describes the chemical degradation of the battery with age. Calculate the voltage across a 14 Ω lamp (the load) when the battery is new. Express the answer with one decimal place.
Rt = 1.7 + 14 = 15.7 Ohms.
I = E/Rt = 8/15.7 = 0.51A,
V = IRl = 0.51 * 14 = 7.1 Volts across
lamp.
To calculate the voltage across the 14 Ω lamp when the battery is new, we need to use Ohm's Law, which states that the voltage across a resistor is equal to the current flowing through it multiplied by its resistance.
First, let's calculate the total resistance in the circuit when the battery is new. Since the internal resistance of the battery is 1.7 Ω and it is connected in series with the 14 Ω lamp, the total resistance (Rt) is given by:
Rt = 1.7 Ω + 14 Ω
Rt = 15.7 Ω
Next, we need to calculate the current flowing through the circuit when the battery is new. We can use Ohm's Law again, this time rearranging the equation to solve for current (I):
I = V / Rt
Where:
I - Current (in Amperes)
V - Voltage across the circuit (in Volts)
Rt - Total resistance (in Ohms)
We are given that the battery voltage is 8 V, so we can substitute the values into the equation:
I = 8 V / 15.7 Ω
I ≈ 0.5102 A (rounded to four decimal places)
Finally, we can calculate the voltage across the 14 Ω lamp using Ohm's Law:
Vlamp = I x Rlamp
Where:
Vlamp - Voltage across the lamp (in Volts)
I - Current flowing through the circuit (in Amperes)
Rlamp - Resistance of the lamp (in Ohms)
Substituting the values:
Vlamp = 0.5102 A x 14 Ω
Vlamp = 7.1434 V (rounded to four decimal places)
Therefore, the voltage across the 14 Ω lamp when the battery is new is approximately 7.1 V.