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# In the circuit shown in the figure:A) Power supplied by the battery is $200W$.B) Current flowing the circuit is $5A$.C) Potential difference across $4\Omega$ is equal to the potential difference across $6\Omega$ resistance.D) Current in wire $AB$ is zero.

Last updated date: 24th Jul 2024
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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 $(V)$.

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.