Question

tha rate of growth of current in a L-R circuit is

Answer 1

Switch is now closed and battery E,inductance L and resistance R are now connected in series
Because of self induced emf current will not immediately reach its steady value but grows at a rate depending on inductance and resistance of the c circuit
Let at any instant I be the c current in the circuit increasing from 0 to a maximum value at a rate of increase dI/dt
Now the potential difference across the inductor is
Vop=LdI/dt
and across resistor is
Vpq=IR
Since 
V=Vop+Vpq
so we have, 
 
Thus rate of increase of current would be, 
 
In the beginning at t=0 when circuit is first closed current begins to grow at a rate, 
 
from the above relation we conclude that greater would be the inductance of the inductor, more slowly the current starts to increase.When the current reaches its steady state value I ,the rate of increase of current becomes zero then from equation (7) we have,
0=(V-IR)/L
or,
I=V/R
From this we conclude that final steady state current in the circuit does not depend on self inductance rather it is same as it would be if only resistance is connected to the source
Now we will obtain the relation of current as a function of time Again consider equation (6) and rearrange it so that
 
let V/R=Imax ,the maximum current in the circuit .so we have

 
Integrating equation (8) on both sides we have

 
where C is a constant and is evaluated by the value for current at t=0 which is i=0
so,
C=-ln(V/R)=-ln Imax
putting this value of C in equation (9) we get, 
 
This equation shows the exponential increase of current in the circuit with the passage of time

Figure below shows the plot of current versus time


 

If we put t=τL=L/R is equation 10 then, 
 
Hence, the time in which the current in the circuit increases from zero to 63% of the maximum value of Imax is called the constant or the decay constant of the circuit.For LR circuit, decay constant is,
τL=L/R                                 ---(11)
Again from equation (8), 
 
This suggests that rate of change current per sec depends on time constant.Higher is the value of decay constant ,lower will be the rate of change of current and vice versa.(B) Decay of currentWhen the switch S is thrown down to b as shown below in the figure ,the L-R circuit is again closed and battery is cut off


 

In this condition the current in the circuit begins to decay
Again from equation (8) since V=0 this time, so the equation for decay is

 
Integrating on both sides

 
In this case initially at time t=0 current I=Imax so 
C1=ln I0 
Putting this value of C1 in equation (12)

 
Hence current decreases exponentially with time in the circuit in accordance with the above equation after the battery are cutoff from the circuit.
Figure below shows the graph between current and time


 

If in equation (13) 
t=τL=L/R 
then
I=Imaxe-1=.37Imax
hence the time in which the current decrease from the maximum value to 37% of the maximum value Imax is called the time constant of the circuit
From equation (13) it is clear that when R is large ,current in the L-R circuit will decrease rapidly and there is a chance of production of spark
To avoid this situation L is kept large enough to make L/R large so that current can decrease slowly
For large time constant the decay is slow and for small time constant the decay is fast

Answer 2

Answer:

1 s

In one time constant 63% change occurs in the value of current. The 63% of maximum current is 1.26 A. If is obvious from graph, the current 1.26 A corresponds to time which is slightly greater than 0.9s. Hence, option having 1 s is most appropriate

Explanation:

1 s

In one time constant 63% change occurs in the value of current. The 63% of maximum current is 1.26 A. If is obvious from graph, the current 1.26 A corresponds to time which is slightly greater than 0.9s. Hence, option having 1 s is most appropriate