Question

The time taken for the magnetic energy to reach $25\%$ of its maximum value, when a solenoid of resistance $R$, inductance $L$ is connected to a battery, is

• A. $\frac{L}{R}\ln \left(2\right)$
• B. $\frac{L}{R}\ln \left(10\right)$
• C. Infinite
• D. $\frac{L}{R}\ln \left(5\right)$

Answer 1

The correct option is A

$\frac{L}{R}\ln \left(2\right)$

Step 1: Given data:

A solenoid of resistance $R$

Inductance $L$

Let $\text{'}t\text{'}$ be the time taken for the magnetic energy to reach $25\%$ of its maximum value

Step 2: Formula used:

Maximum magnetic energy will be given by-

$U_o=\frac{1}{2}L{I_o}^2\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Where, $I_o$ is the maximum current flow through $L-R$ circuit.

From current-time relations, we know that,

$I=I_o\left(1-e^{-\frac{Rt}{L}}\right)\dots \dots \dots \dots \dots \dots \dots \left(2\right)$

Which states that the growth of the current is exponential.

Where, $I$ is the current flowing through $L-R$ circuit at any instant $\text{'}t\text{'}$.

Step 3: Calculation of current:

Let $I$ be the current flowing when the magnetic energy reaches to $25\%$ of its maximum value.

The obtained magnetic energy will be, $U=\frac{1}{2}L{I}^2\dots \dots \dots \dots \dots \dots \dots \left(3\right)$

As it is given that,

$U=\frac{25}{100}{U}_o$

$U=\frac{U_o}{4}$

From equations (1) and (3)-

$$\begin{gathered} \frac{1}{2}L{I}^2=\frac{1}{4}\left(\frac{1}{2}L{I_o}^2\right) \\ I=\frac{I_o}{2}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(4\right) \end{gathered}$$

Thus the current in the $L-R$ circuit when the magnetic energy reached $25\%$ of its maximum value is $I=\frac{I_o}{2}$.

Step 4: Calculation of time taken:

From equation (2), the current flows through $L-R$ circuit at any instant $\text{'}t\text{'}$ is-

$I=I_o\left(1-e^{-\frac{Rt}{L}}\right)$

From equation (4), putting the value of $I$

$$\begin{gathered} \frac{I_o}{2}=I_o\left(1-e^{-\frac{Rt}{L}}\right) \\ \frac{1}{2}=1-e^{-\frac{Rt}{L}} \\ {e}^{-\frac{Rt}{L}}=\frac{1}{2} \end{gathered}$$

Taking log on both sides,

$$\begin{gathered} lo{g}_e\left(\frac{1}{2}\right)=lo{g}_e\left(e^{-\frac{Rt}{L}}\right) \\ \ln \left(\frac{1}{2}\right)=-\frac{Rt}{L} \\ t=-\frac{L}{R}\ln \left(\frac{1}{2}\right) \\ t=\frac{L}{R}\ln \left(2\right) \end{gathered}$$ from the property rule of the natural logarithm $\ln \left(e^x\right)=x\&\ln \left(x\right)=-\ln \left(\frac{1}{x}\right)$

Thus the time taken for the magnetic energy to reach $25\%$ of its maximum value is $\frac{L}{R}\ln \left(2\right)$.

Hence, option (A) is the correct answer.