Grade 12Electric Current
Match List I with List II in the codes given below the Lists.
| List-I (R-C circuit) | List-II (Time constant) | ||
| (P) | (1) | ||
| (Q) | (2) | RC | |
| (R) | (3) | 2RC | |
| (S) | (4) |
Codes :
| P | Q | R | S | |
| (A) | 1 | 2 | 4 | 3 |
| (B) | 1 | 2 | 4 | 2 |
| (C) | 2 | 1 | 3 | 4 |
| (D) | 2 | 1 | 2 | 4 |
- A
- B
- C
- D
1 Answer
vinodh kumar
The correct answer is option B
I will elaborate the procedure below in step by step:::
Step 1 : If there is any parallel or series combinations of Capacitors are present , then reduce them to single equivalent capacitor. let it be Ceq
In JEE syllabus, These type of questions will contain only one capacitor after step 1 is done. These are called as Single Time constant Circuits. As they have only one Time constant (meaning that it has only one equivalent capacitor)
Step 2 : Make Energy source to Zero ( that is volatge source has to short circuited and current source has to be open circuited )
Step 3 : Now take out capacitor from the circuit and find the equivalent resistance across those two terminals from which you removed the capacitor
Let this Equivalent Resistance be Req
so the time constant T is given as T = Ceq Req
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