https://file.aiquickdraw.com/mathbot/file/1726476731393dechbq46.jpg

To find the voltage across the inductor \( v(t) \), we need to analyze the circuit shown in the image. The circuit consists of two current sources with exponential functions, resistors, and an inductor.

### Step-by-step Analysis:

#### 1. **Write the Loop and Node Equations:**
   The two current sources are driving currents \( e^{at} \) and \( e^{bt} \) through the resistors and inductor. Let's start by considering the following:

   - The left branch has a 10-ohm resistor and the current source \( e^{at} \).
   - The middle branch consists of the inductor with inductance \( 1 \, \text{H} \), where the voltage across the inductor is denoted as \( v(t) \).
   - The right branch has a 10-ohm resistor and the current source \( e^{bt} \).

#### 2. **Voltage-Current Relationship for the Inductor:**
   For the inductor, the voltage is related to the current by the equation:
   \[
   v(t) = L \frac{d i_L(t)}{dt}
   \]
   where \( L = 1 \, \text{H} \) and \( i_L(t) \) is the current through the inductor.

#### 3. **Current at the Node (Middle Branch):**
   At the node where the inductor is connected, Kirchhoff's Current Law (KCL) gives:
   \[
   i_L(t) = i_1(t) + i_2(t)
   \]
   where:
   - \( i_1(t) = \frac{e^{at}}{10} \) (current from the left branch),
   - \( i_2(t) = \frac{e^{bt}}{10} \) (current from the right branch).

#### 4. **Calculate the Total Current \( i_L(t) \):**
   Using the above, the total current through the inductor is:
   \[
   i_L(t) = \frac{e^{at}}{10} + \frac{e^{bt}}{10}
   \]
   Simplifying:
   \[
   i_L(t) = \frac{e^{at} + e^{bt}}{10}
   \]

#### 5. **Determine the Voltage \( v(t) \):**
   Now, using the voltage-current relationship for the inductor:
   \[
   v(t) = L \frac{d i_L(t)}{dt}
   \]
   Substituting \( L = 1 \):
   \[
   v(t) = \frac{d}{dt} \left( \frac{e^{at} + e^{bt}}{10} \right)
   \]
   Differentiating:
   \[
   v(t) = \frac{1}{10} \left( a e^{at} + b e^{bt} \right)
   \]

Thus, the voltage across the inductor is:
\[
v(t) = \frac{1}{10} \left( a e^{at} + b e^{bt} \right)
\]

### Conclusion:
The voltage across the inductor is \( v(t) = \frac{1}{10} \left( a e^{at} + b e^{bt} \right) \).

Would you like more details on any part of this solution? Here are some related questions you may explore:

1. What are the general properties of inductors in a circuit?
2. How do we use Kirchhoff's Laws to solve for voltages and currents in complex circuits?
3. What happens if \( a = b \) in the given exponential terms?
4. How does the value of the inductance affect the voltage across it?
5. How can we compute the energy stored in the inductor?

**Tip:** In circuits with inductors, always pay attention to the time-varying nature of the current, as the voltage across an inductor depends on the rate of change of current!

In the given figure, find the voltage across the inductor.

This detailed solution explains how to find the voltage across an inductor in a circuit with exponential current sources. The analysis includes applying Kirchhoff's Current Law and differentiation, making it ideal for undergraduate electrical engineering students.

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