A lead-acid battery of a car has an e.m.f of 12V. If the internal resistance of the battery is 0.5ohm, the maximum current that can be drawn from the battery will be:
$
{\text{A}}{\text{. 30A}} \\
{\text{B}}{\text{. 20A}} \\
{\text{C}}{\text{. 6A}} \\
{\text{D}}{\text{. 24A}} \\
$
Answer
617.1k+ views
Hint: Every battery offers some internal resistance to the flow of current through it which is due to the electrolyte in the battery. A small voltage drop occurs inside the battery and potential drop across the battery is less than the e.m.f. of the battery.
Detailed step by step solution:
The internal resistance is the resistance which is present within the battery that tries to resist the current flow when connected to a circuit. As a result of this resistance, a voltage drop occurs when current flows through the battery. This resistance in the battery is due to the electrolyte and electrodes which are the part of the battery.
This resistance due to the electrolyte and electrodes present in the battery is called internal resistance of the battery which causes a small voltage drop inside the battery.
As we know that the potential difference across a battery is given by the following expression:
\[V = \varepsilon - Ir\]
Here V represents the potential difference across the battery, $\varepsilon $ is the e.m.f. produced by the battery while r is the internal resistance of the battery and I is the amount of current that can be drawn from the battery.
Now, for the given battery if we consider that it is not yet connected to the car then V = 0. Then we have
$\varepsilon - Ir = 0$
We are already given the values of e.m.f. and the internal resistance of the battery, therefore, we have
$I = \dfrac{\varepsilon }{r} = \dfrac{{12}}{{0.5}} = 24A$
Hence, the correct answer is option D.
Note: The obtained amount of current is the maximum amount of current which can be drawn from the battery. The current may be reduced further due to the resistance of the wires used to connect the circuit. The voltage drop is very small so it can be neglected.
Detailed step by step solution:
The internal resistance is the resistance which is present within the battery that tries to resist the current flow when connected to a circuit. As a result of this resistance, a voltage drop occurs when current flows through the battery. This resistance in the battery is due to the electrolyte and electrodes which are the part of the battery.
This resistance due to the electrolyte and electrodes present in the battery is called internal resistance of the battery which causes a small voltage drop inside the battery.
As we know that the potential difference across a battery is given by the following expression:
\[V = \varepsilon - Ir\]
Here V represents the potential difference across the battery, $\varepsilon $ is the e.m.f. produced by the battery while r is the internal resistance of the battery and I is the amount of current that can be drawn from the battery.
Now, for the given battery if we consider that it is not yet connected to the car then V = 0. Then we have
$\varepsilon - Ir = 0$
We are already given the values of e.m.f. and the internal resistance of the battery, therefore, we have
$I = \dfrac{\varepsilon }{r} = \dfrac{{12}}{{0.5}} = 24A$
Hence, the correct answer is option D.
Note: The obtained amount of current is the maximum amount of current which can be drawn from the battery. The current may be reduced further due to the resistance of the wires used to connect the circuit. The voltage drop is very small so it can be neglected.
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