
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.
Answer
218.4k+ views
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.
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.
Recently Updated Pages
Young’s Double Slit Experiment Derivation Explained

Wheatstone Bridge Explained: Working, Formula & Uses

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

