Question

A coil of inductance 1 H and resistance 100 Ω is connected to a battery of 6 V. Determine approximately :
(a) The time elapsed before the current acquires half of its steady – state value.
(b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on.
(Given \(ln2=0.693\), e^–3/2 = 0.25)

• A. t = 10 ms; U = 2 mJ
• B. t = 10 ms; U = 1 mJ
• C. t = 7 ms; U = 1 mJ
• D. t = 7 ms; U = 2 mJ

Answer 1

The Correct Option is C

To solve this problem, we need to address two parts: the time elapsed before the current reaches half of its steady-state value and the energy stored in the magnetic field after a certain time. 

1. Steady-state current calculation:
   • The steady-state current, \(I_0\), can be found using Ohm's Law: \(I_0 = \frac{V}{R}\).
   • Given that the voltage \(V = 6 \, \text{V}\) and resistance \(R = 100 \, \Omega\), we have: \(I_0 = \frac{6}{100} = 0.06 \, \text{A}\).
2. Time for current to reach half of the steady-state value:
   • The current \(I(t)\) at time \(t\) is given by the formula: \(I(t) = I_0(1-e^{-\frac{t}{\tau}})\),
   • where \(\tau = \frac{L}{R}\) is the time constant. 
     For \(L = 1 \, \text{H}\) and \(R = 100 \, \Omega\), \(\tau = \frac{1}{100} = 0.01 \, \text{s} = 10 \, \text{ms}\).
   • For the current to be half its steady-state value: \(\frac{I_0}{2} = I_0(1-e^{-\frac{t}{\tau}})\),
   • This simplifies to: \(e^{-\frac{t}{\tau}} = \frac{1}{2}\).
   • Taking the natural log on both sides, we get: \(-\frac{t}{\tau} = \ln(\frac{1}{2}) = -\ln(2) = -0.693\).
   • Substituting the values, we get: \(-\frac{t}{10 \, \text{ms}} = -0.693 \Rightarrow t \approx 6.93 \, \text{ms} \approx 7 \, \text{ms}\).
3. Energy stored in the magnetic field after 15 ms:
   • The energy stored in the magnetic field, \(U(t)\), is given by: \(U(t) = \frac{1}{2}LI(t)^2\).
   • Let's first find \(I(15 \, \text{ms})\): \(I(t) = 0.06(1-e^{-\frac{t}{\tau}})\) 
     For \(t = 15 \, \text{ms}\), \(\frac{t}{\tau} = \frac{15}{10} = 1.5\),
   • thus \(I(15 \, \text{ms}) = 0.06(1-e^{-1.5}) \approx 0.06(1-0.25) = 0.045 \, \text{A}\).
   • Now, substitute into the formula for energy: \(U(t) = \frac{1}{2} \times 1 \, \text{H} \times (0.045 \, \text{A})^2 = \frac{1}{2} \times 0.002025 \, \text{J} = 0.0010125 \, \text{J}\),
   • which is approximately \(1 \, \text{mJ}\).
4. Conclusion:
   • Therefore, the time elapsed before the current acquires half of its steady-state value is approximately 7 ms, and the energy stored after 15 ms is 1 mJ.

Thus, the correct answer is: t = 7 ms; U = 1 mJ.

Answer 2

\(i(t) = \frac{V}{R} \left(1 - e^{-\frac{Rt}{L}}\right)\) \(⋯(1)\)
\(\frac{L}{R} = \frac{1}{100s}\)

\(⇒\) \(\frac{L}{R} = 10 \ \text{ms} \quad \dots (2)\)

\(\frac{V}{2R} = \frac{V}{R} \left(1 - e^{-\frac{Rt}{L}}\right)\)

\(⇒\) \(e^{-\frac{Rt}{L}} = \frac{1}{2}\)

\(⇒\)\(t = \frac{L}{R} \ln 2 = 6.93 \ \text{ms}\)

\(U = \frac{1}{2}Li^2\)

\(=\frac{1}{2} \left[1 - e^{-\frac{15}{10}}\right]^2 \left(\frac{6}{100}\right)^2\)
\(=\frac{1}{2}[1 - 0.25]^2 \times 36 \times 10^{-4}\)
\(= 1 \)mJ
So, the correct option is (C): t = 7 ms; U = 1 mJ