# A combination of secondary cells has an equivalent EMF of 2 V, internal resistance of 400 ohms. Maximum current that can be drawn is: A) 0.005 A B) 0.0005 A C) 0.010 A D) 0.010 A

## 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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