A series LR circuit contains an emf source of 14 V having no internal resistance

     A series LR circuit contains an emf source of 14 V having no internal resistance, a resistor, a 34 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? a. 1.5 ? b. 1.9 ? c. 5.0 ? d. 14 ?

The Correct Answer and Explanation is :

To solve this problem, we use the equation for the voltage across the inductor in an LR circuit as a function of time:  

\[
V_L(t) = \mathcal{E}(1 - e^{-t/\tau}),
\]

where:  
- \(\mathcal{E} = 14 \, \text{V}\) (emf of the source),  
- \(\tau = L/R\) is the time constant,  
- \(L = 34 \, \text{H}\) is the inductance of the inductor,  
- \(R\) is the resistance we need to find,  
- \(t = 4.0 \, \text{s}\) is the given time.  

The emf across the inductor is 80% of its maximum value (\(\mathcal{E}\)), so:  
\[
V_L(t) = 0.8\mathcal{E} = 0.8 \times 14 = 11.2 \, \text{V}.
\]

Substituting into the equation for \(V_L(t)\):  
\[
11.2 = 14(1 - e^{-t/\tau}).
\]

Divide through by 14:  
\[
\frac{11.2}{14} = 1 - e^{-t/\tau}.
\]

Simplify:  
\[
0.8 = 1 - e^{-t/\tau}.
\]

Rearrange for \(e^{-t/\tau}\):  
\[
e^{-t/\tau} = 0.2.
\]

Take the natural logarithm of both sides:  
\[
-\frac{t}{\tau} = \ln(0.2).
\]

Substitute \(t = 4.0 \, \text{s}\):  
\[
-\frac{4.0}{\tau} = \ln(0.2).
\]

Solve for \(\tau\):  
\[
\tau = \frac{4.0}{-\ln(0.2)}.
\]

Numerically:  
\[
\ln(0.2) \approx -1.609, \quad \tau = \frac{4.0}{1.609} \approx 2.49 \, \text{s}.
\]

The time constant \(\tau\) is also defined as \(\tau = L/R\). Substituting \(\tau = 2.49 \, \text{s}\) and \(L = 34 \, \text{H}\):  
\[
R = \frac{L}{\tau} = \frac{34}{2.49} \approx 13.65 \, \Omega.
\]

The closest option is **d. 14 \(\Omega\)**.  

### Explanation
An LR circuit's time constant (\(\tau\)) determines the rate of change of current and voltage across components. The time constant is influenced by the inductance \(L\) and resistance \(R\). Here, we use the given percentage of maximum voltage (80%) to calculate \(R\), solving through exponential decay equations. This approach is foundational in electrical engineering to analyze transient responses in circuits.
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