A battery of 15volts and internal resistance 3ohms is connected to a resistor of resistance 20ohms calculate both terminal potential difference & Lost volts of the battery
To calculate the terminal potential difference of the battery, we need to consider the voltage drop across its internal resistance.
The total resistance in the circuit is given by the sum of the resistance of the battery (3 ohms) and the external resistor (20 ohms): 3 ohms + 20 ohms = 23 ohms.
Using Ohm's law, we can calculate the current flowing through the circuit: I = V/R, where V is the terminal potential difference of the battery. Rearranging this equation, we have: V = IR.
Substituting the values we have: V = (15 volts)⋅(23 ohms) = 345 volts.
Therefore, the terminal potential difference of the battery is 345 volts.
The lost volts of the battery can be calculated by subtracting the terminal potential difference from the emf (electromotive force) of the battery.
The emf of the battery is given as 15 volts.
Therefore, the lost volts of the battery = emf - terminal potential difference = 15 volts - 345 volts = -330 volts.
Note: The lost volts of the battery indicates that there is a drop in potential difference due to the internal resistance of the battery.
To calculate the terminal potential difference of the battery, we need to take into account its internal resistance and the load resistance (in this case, the resistor of 20 ohms).
The formula for terminal potential difference (Vt) is given by:
Vt = EMF - (I * r)
Where:
EMF is the electromotive force of the battery (given as 15 volts),
I is the current flowing through the circuit, and
r is the internal resistance of the battery (given as 3 ohms).
To find the current, we can use Ohm's Law:
I = V / R
Where:
V is the voltage across the load resistor, and
R is the resistance of the load resistor.
In this case, V is equal to the terminal potential difference (Vt) and R is equal to the resistance of the load resistor (20 ohms).
Let's calculate the current first:
I = Vt / R
I = (15V - (I * 3Ω)) / 20Ω
Now we can solve this equation to find the current:
20I = 15V - 3I
23I = 15V
I = 15V / 23
I ≈ 0.6522 A
Now we can substitute this value for current back into the equation for terminal potential difference:
Vt = 15V - (0.6522A * 3Ω)
Vt = 15V - 1.9566V
Vt ≈ 13.0434 V
Therefore, the terminal potential difference of the battery is approximately 13.0434 volts.
To calculate the lost volts (VL), we can subtract the terminal potential difference (Vt) from the electromotive force (EMF) of the battery:
VL = EMF - Vt
VL = 15V - 13.0434V
VL ≈ 1.9566 V
Therefore, the lost volts of the battery is approximately 1.9566 volts.