In an LR circuit,the current increases to one | vedclass

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

Question

(C) The growth of current in an $LR$ circuit is given by $I = I_0 (1 - e^{-t/	au})$,where $	au$ is the time constant.Given that at $t = 4 \, sec$,$I

Vedclass · Accepted Answer

$\frac{4}{\log_e (4/3)}$

Vedclass · Answer

(C) The growth of current in an $LR$ circuit is given by $I = I_0 (1 - e^{-t/	au})$,where $	au$ is the time constant.Given that at $t = 4 \, sec$,$I = \frac{I_0}{4}$.Substituting these values: $\frac{I_0}{4} = I_0 (1 - e^{-4/	au})$.$\frac{1}{4} = 1 - e^{-4/	au} \implies e^{-4/	au} = 1 - \frac{1}{4} = \frac{3}{4}$.Taking the natural logarithm on both sides: $-\frac{4}{	au} = \log_e (3/4)$.$-\frac{4}{	au} = \log_e 3 - \log_e 4$.$\frac{4}{	au} = \log_e 4 - \log_e 3 = \log_e (4/3)$.Therefore,$	au = \frac{4}{\log_e (4/3)} \, sec$.

In an $LR$ circuit,the current increases to one-fourth of its maximum value in $4 \, sec$. What is the time constant of the circuit?

Similar Questions

Two figures are shown as Fig. $A$ and Fig. $B$. The time constant of Fig. $A$ is $\tau_{A}$ and the time constant of Fig. $B$ is $\tau_{B}$. Then:

$A$ bulb of $100\, W$ is connected in parallel with an ideal inductance of $1\, H$. This arrangement is connected to a $90\, V$ battery through a switch. On pressing the switch,the

In the given $L-R$ circuit,which of the following statements is correct?

The ratio of the time constant in charging and discharging in the circuit shown in the figure is

If $R$ and $L$ denote resistance and inductance respectively,which of the following has the dimension of time?

Vedclass Test Series

Exam Paper Generator

Online Exam Module