Typically , atom gain or lose electrons to achieve A stable electron configuration

The formation of an ionic bond involves : Transfer of electrons
In the formation of ionic bond, the electron is closer to being stolen , not shared

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At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 43.4 m

Sauron [17]

Answer:30.08 ms

Explanation:

Given

time Constant $\tau =43.4 ms =\frac{L}{R}$

Also rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in inductor's magnetic Field

Energy stored in Inductor is $U_L=\frac{1}{2}Li^2$

rate of Energy storing $\frac{dU_L}{dt}=\frac{1}{2}L\cdot 2i\times \frac{di}{dt}----1$

Rate of Energy dissipation from resistor i.e. Power is given by

$\frac{dU_R}{dt}=i^2R-----2$

Equating 1 and 2

$Li\cdot \frac{di}{dt}=i^2R$

$L(\frac{di}{dt})=R(i)-----3$

i is given by $i=\frac{V}{R}(1-e^{-\frac{t}{\tau }})$

$\frac{di}{dt}=\frac{V}{L}e^{-\frac{t}{\tau }}$

substitute the value of $\frac{di}{dt}$ in 3

$L(\frac{V}{L}e^{-\frac{t}{\tau }})=R\cdot \frac{V}{R}(1-e^{-\frac{t}{\tau }})$

$e^{-\frac{t}{\tau }}=1-e^{-\frac{t}{\tau }}$

$2e^{-\frac{t}{\tau }}=1$

$e^{-\frac{t}{\tau }}=0.5$

$e^{-\frac{t}{43.4\times 10^{-3}}}=0.5$

$\frac{t}{43.4\times 10^{-3}}=0.693$

$t=30.08 ms$

4 0

3 years ago