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**Hint:**After determining the circuital validity of the $4\Omega $ and $6\Omega $ resistor arrangement, we will determine the current in the circuit. That value will help us determine the power supplied by it.

**Formula Used:**

Current supplied by the battery: $I = \dfrac{V}{R}$.

Where $V$ is the voltage supplied and is expressed in Volt $(V)$, $R$ is the resistance value and is expressed in Ohms $(\Omega )$ and $I$ is the current value of the battery and is expressed in Ampere $(A)$.

Power supplied by the battery $P = EI$

Where $P$ is the power valueof the battery and is expressed in Watts $(W)$ and $E$ is the emf value of the cell and is expressed in Volt $(V)$.

**Complete step by step answer:**

The arrangement of the $4\Omega ,6\Omega $ is short circuited due to the negligibly low equivalent resistance value of its arrangement.

This means that the voltage difference across its end is zero. That is the potential difference across $4\Omega ,6\Omega $ is the same which is equal to $0$. Which means that no current passes through it.

But, current passes through the internally connected wire. In this case, this wire is $AB$. The value of this current is equal to the current passing through the entire circuit via the battery.

Therefore, the current passing through the battery we use the expression $I = \dfrac{V}{R}$.

Substituting the values we get,

$

I = \dfrac{{20}}{2} \\

\Rightarrow I = 10A \\

$

As we have the current value through the battery we will be able to calculate the power supplied by the same by substituting the values of $E$ and $I$ in $P = EI$.

We get,

$

P = EI = 20 \times 10 \\

\Rightarrow P = 200W \\

$

**In conclusion, the correct options are A and C.**

**Note:**The resistance values of $4\Omega $ and $6\Omega $ are not to be considered while calculating the current in the battery. Also, it is a short circuited arrangement but current passes through the middle wire.

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