Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

In the circuit shown below, all the inductors (assumed ideal) and resistors are identical. The current through the resistance on the right is I after the key K has been switched on for a long time. The currents through the three resistors (in order, from left to right) immediately after the key is switched off are –
A.) 2I upwards, I downwards and I downwards
B.) 2I downwards, I downward and I downwards.
C.) I downwards, I downward and I downwards.
D.) O, I downward and I downwards.

Answer
VerifiedVerified
598.8k+ views
Hint: Consider two cases, in the first case consider that circuit is on. In second when the inductor gets fully charged, it acts as a short circuit. Again divide the current in each branch and devise such that the current loop must be satisfied. While solving the question, notice the direction of current in each device.

Complete step by step solution:
Diagram:
seo images


In the question, it is given that all the inductors and resistors are identical.

i.e. \[{{R}_{1}}={{R}_{2}}={{R}_{3}}\text{ and }{{L}_{1}}={{L}_{2}}\]

Condition is given that, if a switch is on for a long time then current will pass through the right side resistor. If a switch is on for a long time then, the inductor will get charges and our circuit will act like a short circuit. Therefore current passing through \[{{L}_{2}}\]and \[{{R}_{3}}\] will also be I. If the current in \[{{R}_{3}}\] is I then current in \[{{R}_{2}}\] will also be I. if current in \[{{L}_{2}}\] and \[{{R}_{2}}\] is I then it must be split at junction 2 after passing through\[{{L}_{1}}\]. Therefore current in \[{{L}_{1}}\] is 2I.

Now consider junction 3:

Current at junction 3 is coming from \[{{R}_{2}}\] and \[{{R}_{3}}\]. Therefore current leaving from junction 3 is 2I.

Now consider junction 4:

Current 2I will again divide into I in resistance \[{{R}_{1}}\] and E (cell). Here the direction of the current will be in an upward direction. Whereas the direction of current in \[{{R}_{3}}\] and \[{{R}_{2}}\]is downward.

So up till now, we are done with dividing current into the branches. This case happens when the switch is on.

In the question, it asks the currents through the three resistors (in order, from left to right) immediately after the key is switched off are:

When the switch is off, it acts as a short circuit therefore the circuit on the left side will get short. Therefore no current will be split at junction 4 and current in \[{{R}_{1}}\] is 2I.
So options A is correct since it is matching to our circuit direction.

Answer- (A)

Note: Note that current coming to a junction and leaving from the junction must be zero. So while splitting the current to branches and devices, the current loop must be satisfied. When current in devices exceeds the limit then current is said to be short. If the value of all resistance is the same then the current flowing through it will be the same if it is parallel to each other.