A Storage battery of a car has an EMF of 12 volt if the internal resis...
Understanding the Problem:
To find the maximum current that can be drawn from a car battery, we need to consider the electromotive force (EMF) and the internal resistance of the battery. The EMF represents the potential difference across the terminals of the battery when there is no current flowing, while the internal resistance represents the resistance within the battery itself.
Given Data:
- EMF of the battery: 12 volts
- Internal resistance of the battery: 0.4 ohms
Calculating the Maximum Current:
The maximum current that can be drawn from the battery can be calculated using Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) across the circuit divided by the resistance (R) of the circuit.
Using Ohm's Law, we can determine the maximum current (I_max) as follows:
I_max = EMF / (Internal resistance + Load resistance)
In this case, the load resistance represents the resistance of the external circuit that is connected to the battery.
Example:
Let's assume the load resistance is 1 ohm. We can calculate the maximum current as follows:
I_max = 12 volts / (0.4 ohms + 1 ohm)
I_max = 12 volts / 1.4 ohms
I_max ≈ 8.57 amps
Therefore, the maximum current that can be drawn from the battery is approximately 8.57 amps when the load resistance is 1 ohm.
Conclusion:
The maximum current that can be drawn from a car battery depends on the EMF of the battery and the internal resistance of the battery. By using Ohm's Law, we can calculate the maximum current by dividing the EMF by the sum of the internal resistance and the load resistance. It is important to consider the load resistance when determining the maximum current, as a lower load resistance will result in a higher current draw.
A Storage battery of a car has an EMF of 12 volt if the internal resis...
Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,
E = Ir
I = E / r
= 12 / 0.4
= 30 A
The maximum current drawn from the given battery is 30 A.