Question

A combination of secondary cells has an equivalent EMF of $2 \mathrm{~V}$, internal resistance of $400 \Omega$. Maximum current that can be drawn is:

A) $0.005 \mathrm{~A}$

B) $0.0005 \mathrm{~A}$

C) $0.010 \mathrm{~A}$

D) $0.010 \mathrm{~A}$

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Answer

The correct answer is A) $0.005 \mathrm{~A}$.

To understand how to derive this, consider the key formula needed to find the current through the circuit:

$$ i = \frac{V}{R} $$

where $V$ is the voltage (or EMF) of the battery and $R$ is the internal resistance. In this problem, you are provided with an EMF of $2 \mathrm{V}$ and an internal resistance of $400 \Omega$.

To draw the maximum current from the battery, you need to minimize the external resistance connected across it, effectively making it $0 \Omega$. This scenario is known as short-circuiting the battery.

Applying the formula:

$$ i = \frac{2 \mathrm{V}}{400 \Omega} = 0.005 \mathrm{A} $$

This calculation confirms that the maximum current that can be drawn from this configuration is $0.005 \mathrm{~A}$. Thus, the correct answer is A) $0.005 \mathrm{~A}$.


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