Solveeit Logo

Question

Question: In the circuit shown in the figure: ![](https://www.vedantu.com/question-sets/93c13c18-1290-442...

In the circuit shown in the figure:

A) Power supplied by the battery is 200W200W.
B) Current flowing the circuit is 5A5A.
C) Potential difference across 4Ω4\Omega is equal to the potential difference across 6Ω6\Omega resistance.
D) Current in wire ABAB is zero.

Explanation

Solution

After determining the circuital validity of the 4Ω4\Omega and 6Ω6\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=VRI = \dfrac{V}{R}.
Where VV is the voltage supplied and is expressed in Volt (V)(V), RR is the resistance value and is expressed in Ohms (Ω)(\Omega ) and II is the current value of the battery and is expressed in Ampere (A)(A).

Power supplied by the battery P=EIP = EI
Where PP is the power valueof the battery and is expressed in Watts (W)(W) and EE is the emf value of the cell and is expressed in Volt (V)(V).

Complete step by step answer:
The arrangement of the 4Ω,6Ω4\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 4Ω,6Ω4\Omega ,6\Omega is the same which is equal to 00. Which means that no current passes through it.
But, current passes through the internally connected wire. In this case, this wire is ABAB. 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=VRI = \dfrac{V}{R}.
Substituting the values we get,
I=202 I=10A  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 EE and II in P=EIP = EI.
We get,
P=EI=20×10 P=200W  P = EI = 20 \times 10 \\\ \Rightarrow P = 200W \\\
In conclusion, the correct options are A and C.

Note: The resistance values of 4Ω4\Omega and 6Ω6\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.