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Rise and Decay of Current In Inductive Circuit

Rise and Decay of Current In Inductive Circuit :- This article explains the rise and decay of current in inductive circuit (often called the rise and decay of current in an inductive circuit) by discussing how an inductor resists sudden changes in current and how the current gradually reaches its steady-state value or falls to zero.

Let a resistance R and an inductor L be connected to a battery of emf ε with the help of a Morse key K as shown in the figure below :-

(a) Rise of Current In an Inductive Circuit

(Rise and Decay of Current In Inductive Circuit)

As soon as the Morse key K is pressed, the resistance R and inductor L get connected to the battery and the current in the R – L circuit starts increasing. Due to self-induction, an electromotive force (e.m.f.) is induced in the inductor L and according to Lenz’s law, this emf opposes the increase in current.

If at any time the value of electric current in the circuit is I and the rate of change of current is dI/dt, then

Potential difference across the resistance R = $\displaystyle IR$

Potential difference across the inductor L = $\displaystyle L\frac{dI}{dt}$

Hence, at some time t,

$\displaystyle \varepsilon =IR+L\frac{dI}{dt}$ …..(1)

When the current in the circuit is maximum, the rate of change of current is zero, i.e., when $\displaystyle \frac{dI}{dt}=0$ , $\displaystyle I={{I}_{0}}$ Therefore, in equation (1), at that time ε = I₀R. From equation (1),

$\displaystyle {{I}_{0}}R=IR+L\frac{dI}{dt}$

$\displaystyle \Rightarrow {{I}_{0}}R-IR=L\frac{dI}{dt}$

$\displaystyle \Rightarrow R\left( {{I}_{0}}-I \right)=L\frac{dI}{dt}$

$\displaystyle \Rightarrow \frac{dI}{\left( {{I}_{0}}-I \right)}=\frac{R}{L}dt$

Integrating both sides within appropriate limits,

$\displaystyle \int\limits_{0}^{I}{\frac{dI}{\left( {{I}_{0}}-I \right)}}=\int\limits_{0}^{\text{t}}{\frac{R}{L}dt}$

$\displaystyle \left( -1 \right)\left[ lo{{g}_{e}}\left( {{I}_{0}}-I \right) \right]_{0}^{I}=\frac{R}{L}\left[ t \right]_{0}^{t}$

$\displaystyle \Rightarrow \left[ lo{{g}_{e}}\left( {{I}_{0}}-I \right) \right]_{0}^{I}=-\frac{R}{L}\left[ t \right]_{0}^{t}$

$\displaystyle \Rightarrow \left[ lo{{g}_{e}}\left( {{I}_{0}}-I \right)-lo{{g}_{e}}\left( {{I}_{0}}-0 \right) \right]=-\frac{R}{L}\left[ t-0 \right]$

$\displaystyle \Rightarrow lo{{g}_{e}}\frac{{{I}_{0}}-I}{{{I}_{0}}}=-\frac{R}{L}t$

$\displaystyle \Rightarrow \frac{{{I}_{0}}-I}{{{I}_{0}}}={{e}^{-\frac{R}{L}t}}$

$\displaystyle \Rightarrow 1-\frac{I}{{{I}_{0}}}={{e}^{-\frac{R}{L}t}}$

$\displaystyle \therefore I={{I}_{0}}\left( 1-{{e}^{-\frac{R}{L}t}} \right)={{I}_{0}}\left( 1-{{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}} \right)$ …..(2)

Where,

$\displaystyle \tau =\frac{L}{R}$

Equation (2) is called the Helmholtz equation for the rise of current in an inductor, and it shows that the current grows exponentially with time in an inductor.

Time Constant

The quantity $\displaystyle \tau =\frac{L}{R}$ is called the time constant of the L-R circuit because $\displaystyle \frac{L}{R}$ has the dimension of time and its value is constant for a given L-R circuit.

Substituting $\displaystyle t=\frac{L}{R}$ in equation (2),

$\displaystyle I={{I}_{0}}\left( 1-{{e}^{-\frac{R}{L}\times \frac{L}{R}}} \right)={{I}_{0}}\left( 1-{{e}^{-1}} \right)={{I}_{0}}\left( 1-\frac{1}{e} \right)={{I}_{0}}\left( 1-\frac{1}{2.718} \right)$

$\displaystyle \Rightarrow I={{I}_{0}}\left( 1-0.368 \right)=0.632{{I}_{0}}$

∴ I = 63.2 % I₀

Therefore, the time constant of an L–R circuit is defined as the time in which the current in the circuit reaches 63.2% of its maximum value.

Again in equation (2) for I = I₀ ,

$\displaystyle {{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}=0$

$\displaystyle \Rightarrow t=\infty$

That is, the current in an L-R circuit takes infinite time to reach its maximum value. However, in practice, the current reaches very close to its maximum value in a time equal to about five times the time constant. The following graph shows the rise of current in an L–R circuit :-

(b) Decay of Current In an Inductive Circuit

(Rise and Decay of Current In Inductive Circuit)

As soon as the Morse key K is released, the battery is disconnected from the circuit and the current begins to decay.

If at any time t the value of current in the circuit is I , then since the battery is disconnected from the circuit, we put ε = 0 in Equation (1).

$\displaystyle IR+L\frac{dI}{dt}=\text{0}$

$\displaystyle \Rightarrow L\frac{dI}{dt}=-IR$

$\displaystyle \Rightarrow \frac{dI}{I}=-\frac{R}{L}dt$

Integrating both sides within appropriate limits,

$\displaystyle \int\limits_{{{I}_{0}}}^{I}{\frac{dI}{I}}=-\frac{R}{L}\int\limits_{0}^{t}{dt}$

$\displaystyle \left[ lo{{g}_{e}}\left( I \right) \right]_{{{I}_{0}}}^{I}=-\frac{R}{L}\left[ t \right]_{0}^{t}$

$\displaystyle \Rightarrow \left[ lo{{g}_{e}}\left( I \right)-lo{{g}_{e}}\left( {{I}_{0}} \right) \right]=-\frac{R}{L}\left[ t-0 \right]$

$\displaystyle \Rightarrow lo{{g}_{e}}\frac{I}{{{I}_{0}}}=-\frac{R}{L}t$

$\displaystyle \Rightarrow \frac{I}{{{I}_{0}}}={{e}^{-\frac{R}{L}t}}$

$\displaystyle \therefore I={{I}_{0}}{{e}^{-\frac{R}{L}t}}={{I}_{0}}{{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}$ …..(3)

Where,

$\displaystyle \tau =\frac{L}{R}$

Equation (3) is called the Helmholtz equation of decay of current in an inductor.

Time Constant

As discussed earlier, the quantity $\displaystyle \tau =\frac{L}{R}$ is called the time constant of the L-R circuit. Substituting $\displaystyle t=\frac{L}{R}$ in Equation (3), we get

$\displaystyle I={{I}_{0}}{{e}^{-1}}=\frac{{{I}_{0}}}{e}=\frac{{{I}_{0}}}{2.718}=0.368{{I}_{0}}$

∴ I = 36.8 % I₀

Therefore, the time constant of an L–R circuit during current decay is the time in which the current in the circuit falls to 36.8% of its maximum value.

Again in equation (3) for I = 0 ,

$\displaystyle I=0$

$\displaystyle \Rightarrow {{I}_{0}}{{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}=0$

$\displaystyle \Rightarrow {{e}^{-{}^{t}\!\!\diagup\!\!{}_{\tau }\;}}=0$

$\displaystyle \Rightarrow t=\infty$

That is, during current decay in an L–R circuit, the current theoretically takes an infinite amount of time to reach zero. The following graph shows the decay of current in an L–R circuit :-

Note :-

If the value of time constant $\displaystyle t=\frac{L}{R}$ is small, then from equation (2) the current takes less time to reach its maximum value (I₀) and from equation (3) the current will also take less time to become zero. Similarly, if the value of time constant $\displaystyle t=\frac{L}{R}$ is large then both the rise and fall of current will take more time.

Therefore, the time constant of an L–R circuit ( $\displaystyle \tau =\frac{L}{R}$ ) is a measure of the rate of rise and decay of current.

Next Topic :- Charging and Discharging of Capacitor

Previous Topic :- Alternating Current

Complete List of Topics :-

Rise and Decay of Current In Inductive Circuit

(a) Rise of Current In an Inductive Circuit

(Rise and Decay of Current In Inductive Circuit)

(b) Decay of Current In an Inductive Circuit

(Rise and Decay of Current In Inductive Circuit)

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