For the given circuit, find the ratio of instantaneous voltage across the inductor when current is \(2\,\text{mA}\) and when current is \(4\,\text{mA}\).
Step 1: Understanding the Concept:
In an RL circuit connected to a DC voltage source, the total voltage provided by the source is distributed between the resistor and the inductor. By applying Kirchhoff's Voltage Law (KVL), we can find the exact voltage drop across the inductor at any instantaneous current.
Step 2: Key Formula or Approach:
KVL Equation for the loop:
\[ V - iR - V_{\text{ind}} = 0 \implies V_{\text{ind}} = V - iR \]
Step 3: Detailed Explanation:
Given source voltage \(V = 3 \, \text{V}\) and resistance \(R = 6 \, \Omega\).
When current \(i = 2 \, \text{mA} = 2 \times 10^{-3} \, \text{A}\):
Voltage drop across resistor: \(V_{R1} = iR = (2 \times 10^{-3}) \times 6 = 0.012 \, \text{V}\)
Voltage across inductor: \(V_{\text{ind1}} = 3 - 0.012 = 2.988 \, \text{V}\)
When current \(i = 4 \, \text{mA} = 4 \times 10^{-3} \, \text{A}\):
Voltage drop across resistor: \(V_{R2} = iR = (4 \times 10^{-3}) \times 6 = 0.024 \, \text{V}\)
Voltage across inductor: \(V_{\text{ind2}} = 3 - 0.024 = 2.976 \, \text{V}\)
Now, calculate the ratio:
\[ \text{Ratio} = \frac{V_{\text{ind1}}}{V_{\text{ind2}}} = \frac{2.988}{2.976} \approx 1.004 \]
Since \(1.004\) is overwhelmingly close to \(1\), the ratio is effectively \(1\).
Step 4: Final Answer:
The ratio is approximately \(1\).
