An inductor coil stores U energy when I curre | vedclass

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

Question

(C) The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$,where $L$ is the inductance and $I$ is the current.From this,we can express t

Vedclass · Accepted Answer

$\frac{2U}{P}$

Vedclass · Answer

(C) The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$,where $L$ is the inductance and $I$ is the current.From this,we can express the inductance as $L = \frac{2U}{I^2}$.The rate of energy dissipation (power loss) in the inductor due to its internal resistance $R$ is given by $P = I^2 R$.From this,we can express the resistance as $R = \frac{P}{I^2}$.The time constant $	au$ of an $LR$ circuit is defined as $	au = \frac{L}{R}$.Substituting the expressions for $L$ and $R$:$	au = \frac{2U / I^2}{P / I^2} = \frac{2U}{P}$.Therefore,the time constant is $\frac{2U}{P}$.

An inductor coil stores $U$ energy when $I$ current is passed through it and dissipates energy at the rate of $P$. The time constant of the circuit,when this coil is connected across a battery of zero internal resistance,is

Similar Questions

In an $L-R$ decay circuit,the initial current at $t = 0$ is $I$. The total charge that has flown through the resistor until the energy in the inductor has reduced to one-fourth of its initial value is:

As shown in the figure,a battery of $emf \; \varepsilon$ is connected to an inductor $L$ and resistance $R$ in series. The switch $S$ is closed at $t=0$. The total charge that flows from the battery between $t=0$ and $t=t_{c}$ ($t_{c}$ is the time constant of the circuit) is

In the figure,a lamp $P$ is in series with an iron-core inductor $L$. When the switch $S$ is closed,the brightness of the lamp rises relatively slowly to its full brightness compared to the case without the inductor. This is due to:

$A$ $50 \, V$ potential difference is suddenly applied to a coil with $L = 5 \times 10^{-3} \, H$ and $R = 180 \, \Omega$. The rate of increase of current after $0.001 \, s$ is ....... $A/s$.

Calculate the current,a long time after the closing of the key $K$ in a circuit containing a battery of $10 \ V$,a resistor of $2 \ \Omega$,and an inductor of $5 \ H$. (in $A$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module