In the circuit shown in the figure,R = sqrt{f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The current in the $LR$ branch at time $t$ is given by $I_L = I_0 (1 - e^{-tR/L})$,where $I_0 = V/R$.The current in the $RC$ branch at time $t$ is

Vedclass · Accepted Answer

$CR \ln 2$

Vedclass · Answer

(B) The current in the $LR$ branch at time $t$ is given by $I_L = I_0 (1 - e^{-tR/L})$,where $I_0 = V/R$.The current in the $RC$ branch at time $t$ is given by $I_C = I_0 e^{-t/RC}$,where $I_0 = V/R$.Equating the two currents: $I_0 (1 - e^{-tR/L}) = I_0 e^{-t/RC}$.$1 - e^{-tR/L} = e^{-t/RC}$.Given $R = \sqrt{\frac{L}{C}}$,we have $R^2 = \frac{L}{C}$,which implies $\frac{L}{R} = RC$.Substituting $\frac{L}{R} = RC$ into the equation: $1 - e^{-t/RC} = e^{-t/RC}$.$1 = 2 e^{-t/RC}$.$e^{t/RC} = 2$.Taking the natural logarithm on both sides: $\frac{t}{RC} = \ln 2$.Therefore,$t = RC \ln 2$.

In the circuit shown in the figure,$R = \sqrt{\frac{L}{C}}$. Switch $S$ is closed at time $t = 0$. The current through $C$ and $L$ would be equal after a time $t$ equal to

Similar Questions

Two identical capacitors $A$ and $B$ are charged to the same potential $V$ and are connected in two circuits at $t = 0,$ as shown in the figure. The charge on the capacitors at time $t = CR$ are respectively:

Find the time constant for the given circuit.

In the circuit shown in the figure,what happens in the steady state?

In an $RC$ circuit while charging,the graph of $\log i$ versus time is as shown by the dotted line in the diagram,where $i$ is the current. When the value of the resistance is doubled,which of the solid curves best represents the variation of $\log i$ versus time?

At steady state,the charge on the capacitor,as shown in the circuit below,is . . . . . . $\mu \text{C}.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module