
Find out the current \[i\] at time \[t=0\] and $t=\infty $ respectively for the circuit that is being represented below.

a. $\dfrac{18}{55}E,\dfrac{5}{18}E$.
b. $\dfrac{5}{18}E,\dfrac{18}{55}E$.
c. $\dfrac{5}{18}E,\dfrac{10}{33}E$.
d. $\dfrac{10}{33}E,\dfrac{5}{18}E$.
Answer
164.1k+ views
Hint: We have to consider the states of inductor at the initial time and the final time. The resistances of the circuit would depend upon the sate of the inductor. Initially the inductor would not work but after infinite time it would work like a wire.
Complete answer:
Here, we are going to provide a complete step by step solution.
We have to consider two parts to get the answer regarding to the solution.
(A Let us consider first when the time $t=0$.
At $t=0$ the circuit at the inductor becomes open. It is being represented below:

Here, the left box and the right box are in parallel connection to each other. Here, ABCF is in series connection and FCDE is also in series connection. But, ABCF is in parallel connection to FCDE.
So, let us find the calculation of the resistance now.
In ABCF we get,
Resistance $=(5+1)\Omega =6\Omega $
In CDFE we get,
Resistance $=(5+4)\Omega =9\Omega $
So, the total resistance $R$ in the circuit is,
$R=\dfrac{6\times 9}{6+9}=\dfrac{54}{15}=\dfrac{18}{5}\Omega $
Hence, the current $i$ in the circuit at time $t=0$ with cell potential $E$ is,
$i=\dfrac{E}{R}=\dfrac{5}{18}E$
(B) Let us consider for time $t=\infty $.
When the time $t=\infty $, then the inductor behaves as an connecting wire.
The diagram is represented below,

The resistance between A and B is in parallel connection.
So, the resistance between A and B $=\dfrac{4\times 1}{4+1}=\dfrac{4}{5}\Omega $
Again, the resistance between B and C is in parallel connection.
So, the resistance between B and C $=\dfrac{5\times 5}{5+5}=\dfrac{5}{2}\Omega $
Hence, the total resistance $R'$ in the circuit $=\dfrac{4}{5}+\dfrac{5}{2}=\dfrac{33}{10}\Omega $.
Now, the current $i$ at time $t=\infty $ is,
$i=\dfrac{E}{R'}=\dfrac{10}{33}E$.
So, the correct option is c. $\dfrac{5}{18}E,\dfrac{10}{33}E$.
Note: According to Newton’s third law of motion, if an object exerts a force on another object, then that object must exert a force of equal magnitude and opposite direction back on the first object. This law of motion signifies a specific symmetry in nature that forces always occur in pairs and one body cannot exert a force on the other body without experiencing a force itself.
Complete answer:
Here, we are going to provide a complete step by step solution.
We have to consider two parts to get the answer regarding to the solution.
(A Let us consider first when the time $t=0$.
At $t=0$ the circuit at the inductor becomes open. It is being represented below:

Here, the left box and the right box are in parallel connection to each other. Here, ABCF is in series connection and FCDE is also in series connection. But, ABCF is in parallel connection to FCDE.
So, let us find the calculation of the resistance now.
In ABCF we get,
Resistance $=(5+1)\Omega =6\Omega $
In CDFE we get,
Resistance $=(5+4)\Omega =9\Omega $
So, the total resistance $R$ in the circuit is,
$R=\dfrac{6\times 9}{6+9}=\dfrac{54}{15}=\dfrac{18}{5}\Omega $
Hence, the current $i$ in the circuit at time $t=0$ with cell potential $E$ is,
$i=\dfrac{E}{R}=\dfrac{5}{18}E$
(B) Let us consider for time $t=\infty $.
When the time $t=\infty $, then the inductor behaves as an connecting wire.
The diagram is represented below,

The resistance between A and B is in parallel connection.
So, the resistance between A and B $=\dfrac{4\times 1}{4+1}=\dfrac{4}{5}\Omega $
Again, the resistance between B and C is in parallel connection.
So, the resistance between B and C $=\dfrac{5\times 5}{5+5}=\dfrac{5}{2}\Omega $
Hence, the total resistance $R'$ in the circuit $=\dfrac{4}{5}+\dfrac{5}{2}=\dfrac{33}{10}\Omega $.
Now, the current $i$ at time $t=\infty $ is,
$i=\dfrac{E}{R'}=\dfrac{10}{33}E$.
So, the correct option is c. $\dfrac{5}{18}E,\dfrac{10}{33}E$.
Note: According to Newton’s third law of motion, if an object exerts a force on another object, then that object must exert a force of equal magnitude and opposite direction back on the first object. This law of motion signifies a specific symmetry in nature that forces always occur in pairs and one body cannot exert a force on the other body without experiencing a force itself.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Charging and Discharging of Capacitor

Instantaneous Velocity - Formula based Examples for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Other Pages
Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Three mediums of refractive indices mu 1mu 0 and mu class 12 physics JEE_Main

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks
