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Question

An emf is 15 V is applied in a circuit coil containing 5 H inductance and 10 &Omega; resistance. The ratio of currents at time $t = \infty $ and t=1 s is

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
**Primary Concept**: Current growth in an RL circuit
**Related Topics**: Inductance, Resistance, Time Constant
**Prerequisites**: Understanding of RL circuits and their behavior, exponential functions
## SOLUTION PATHWAY
**Given Information**:
    * EMF (E) = 15 V
    * Inductance (L) = 5 H
    * Resistance (R) = 10 Ω
    * Time 1 (t1) = 1 s
    * Time 2 (t2) = ∞
**Method & Steps**:
    1. **Find the current at t=∞ (steady state):**
        Reasoning: At t=∞, the inductor acts as a short circuit, and the current reaches its maximum value.
        Formula/Rule:  $I_{\infty} = \frac{E}{R}$
        Working: $I_{\infty} = \frac{15}{10} = 1.5 A$
    2. **Find the current at t=1s:**
        Reasoning: The current in an RL circuit grows exponentially with time.
        Formula/Rule: $I(t) = I_{\infty}(1 - e^{-t/	au})$, where τ is the time constant.
        Working: The time constant is  $	au = \frac{L}{R} = \frac{5}{10} = 0.5 s$. So, $I(1) = 1.5(1 - e^{-1/0.5}) = 1.5(1 - e^{-2})$.
    3. **Calculate the ratio of currents:**
        Reasoning: We need to find the ratio of  $I_{\infty}$ to I(1)
        Working: $\frac{I_{\infty}}{I(1)} = \frac{1.5}{1.5(1 - e^{-2})} = \frac{1}{1 - e^{-2}} = \frac{1}{\frac{e^2 - 1}{e^2}} = \frac{e^2}{e^2-1}$
 ## OPTION ANALYSIS
[A] :x: Incorrect.
[B] ✓: Correct. This matches the derived ratio.
[C] :x: Incorrect.
[D] :x: Incorrect.
 ## LEARNING AIDS
**Common Mistakes**:
    1. **Forgetting the time constant**: Ensure correct calculation of the time constant (τ = L/R).
    2. **Incorrect use of exponential function**:  Pay close attention to the sign of the exponent and how it affects growth or decay.
**Quick Tips**:
    1. Remember that at t=∞, the inductor acts like a short circuit.
    2. The current in an RL circuit grows exponentially towards its steady-state value.

SUBHANI · Answer

$\frac{{{e^{1/2}}}}{{{e^{1/2}} - \;1}}$

SUBHANI · Answer

The correct Option is "$\frac{{{e^2}}}{{{e^2} - 1}}$"

SUBHANI · Answer

$1 - {e^{ - 1}}$

SUBHANI · Answer

${e^{ - 1}}$

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