Question

the time constant of circuit is 10sec and resistance 10 ohm is connected in series in a previous circuit then time constant become 2sec the self inductance of circuit is​

Answer 1

Answer:

Explanation:

Given The time constant of circuit is 10 sec and resistance 10 ohm is connected in series in a previous circuit then time constant become 2 sec the self inductance of circuit is

Time constant = 10 secs , R = 10 ohms

We know that  

Time constant = L / R

Therefore L = time constant x R

                          = 10 x 10  

                         = 100  

Now when T = 2 secs,  

R = 100 / 2 = 50 ohms

Answer 2

Answer:25 hertz

Explanation:

10 = L/R (i)

2= L/R+10 (ii)

2R+20=L

From equation (ii)

10=2R+20/R

=> 8R = 20 or R =5/2

10= 2L/5

=> L=25✓✓✓