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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.

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Last updated date: 24th Feb 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 $(V)$.

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.