Question

An LR circuit with emf ε is connected at t = 0. (a) Find the charge Q which flows through the battery during 0 to t. (b) Calculate the work done by the battery during this period. (c) Find the heat developed during this period. (d) Find the magnetic field energy stored in the circuit at time t. (e) Verify that the results in the three parts above are consistent with energy conservation.

Answer 1

(a) Let the current in the LR circuit be i.

Let the charge flowing through the coil in the infinitesimal time dt be dq.

Now,

\[i = \frac{dq}{dt}\]

∴ dq = idt

The current in the LR circuit after t seconds after connecting the battery is given by

i = i₀ (1 − e^−t/τ)

Here,

i₀ = Steady state current

τ = Time constant = `L/R`

dq = i₀ (1 − e^−tR/L) dt

On integrating both sides, we get

\[Q = \int\limits_0^t dq\]

\[= i_0 \left[ \int\limits_0^t dt - \int\limits_0^t e^{- tR/L} dt \right]\]

\[ = i_0 \left[ t - \left( - \frac{L}{R} \right)( e^{- tR/L} - 1) \right]\]

\[ = i_0 \left[ t - \frac{L}{R}(1 - e^{- tR/L} ) \right]\]

Thus, the charge flowing in the coil in time t is given by

\[Q = \frac{\epsilon}{R}\left[ t - \frac{L}{R}(1 - e^{- tR/L} ) \right]\]

Where `i_0=epsilon/R`

(b) The work done by the battery is given by

W = εQ

From the above expression for the charge in the LR circuit, we have

\[W = \frac{\epsilon^2}{R}\left[ t - \frac{L}{R}(1 - e^{- tR/L} ) \right]\]

(c) The heat developed in time t can be calculated as follows:-

\[H =  \int\limits_0^t  i^2   Rdt\]

\[H = \frac{\epsilon^2}{R^2}R \int\limits_0^t (1 -  e^{- tR/L}  )^2 dt\]

\[H = \frac{\epsilon^2}{R} \int\limits_0^t \left[ (1 + e^{- 2tR/L} ) - 2 e^{- tR/L} \right]dt\]

\[H = \frac{\epsilon^2}{R} \left( t - \frac{L}{2R} e^{- 2tR/L} + \frac{L}{R}2 e^{- tR/L} \right)_0^t \]

\[H = \frac{\epsilon^2}{R}\left( t - \frac{L}{2R} e^{- 2tR/L} + \frac{L}{R}2 e^{- tR/L} \right) - \left( - \frac{L}{2R} + \frac{2L}{R} \right)\]

\[       = \frac{\epsilon^2}{R}\left\{ t - \frac{L}{2R}( x^2 - 4x + 3) \right\}...........\left(x =  e^{- tR/L}\right)\]

(d) The magnetic energy stored in the circuit is given by

\[U = \frac{1}{2}L i^2 \]

\[ \Rightarrow U = \frac{1}{2}L\frac{\epsilon^2}{R^2} \left(1 - e^{- tR/L}\right)^2 \]

\[\Rightarrow U = \frac{L \epsilon^2}{2 R^2}(1 - x )^2\]

(e) Taking the sum of total energy stored in the magnetic field and the heat developed in time t

\[E = \frac{L \epsilon^2}{2 R^2}\left(1 - x\right)^2 + \frac{\epsilon^2}{R}\left\{ t - \frac{L}{2R}\left(x^2 - 4x + 3\right) \right\}\]

\[E = \frac{L \epsilon^2}{2 R^2}\left(1 + x^2 - 2x\right) + \frac{\epsilon^2}{R}t - \frac{L \epsilon^2}{2 R^2}\left(x^2 - 4x + 3\right)\]

\[E = \frac{L \epsilon^2}{2 R^2}\left(2x - 2\right) + \frac{\epsilon^2}{R}t\]

\[E = \frac{L \epsilon^2}{R^2}\left(x - 1\right) + \frac{\epsilon^2}{R}t\]

\[E = \frac{\epsilon^2}{R}\left\{ t - \frac{L}{R}\left(1 - x\right) \right\} = \frac{\epsilon^2}{R}\left\{ t - \frac{L}{R}\left(1 - e^{- tR/L}\right) \right\}\]

The above expression is equal to the energy drawn from the battery. Therefore, the conservation of energy holds good.