Sure! Let’s break down the circuit step by step and solve each part. --- ## **Given:** - The current \( i_a \) is **9.5 mA**. - The circuit contains 1 kΩ, 2 kΩ, 3 kΩ, and 4 kΩ resistors. - You need to find \( i_ \), \( i_g \), and the power delivered by the source. --- ## **Step 1: Understanding the Circuit** ### **Circuit Branches** - The current source \( i_g \) splits into two branches at the top left: - **Branch 1:** \( 1\,k\Omega \) and \( 2\,k\Omega \) in series (top). - **Branch 2:** \( 3\,k\Omega \) and \( 4\,k\Omega \) in series (bottom). - \( i_ \) flows through the **2 kΩ** resistor. - \( i_a \) flows through the **4 kΩ** resistor (given as 9.5 mA). --- ## **Step 2: Find \( i_ \) (Part A)** ### **Current Division** Since the two parallel branches are: - Top: \( 1k + 2k = 3k\Omega \) - Bottom: \( 3k + 4k = 7k\Omega \) The total resistance \(R_{eq}\) is: \[ \frac{1}{R_{eq}} = \frac{1}{3k} + \frac{1}{7k} \] \[ R_{eq} = \frac{3k \times 7k}{3k + 7k} = \frac{21k^2}{10k} = 2.1k\Omega \] The voltage across both branches is the same. Let’s use \(i_a\) to find the voltage: \[ V = i_a \times 4k\Omega = 9.5\,mA \times 4k\Omega = 38\,V \] ### **Now, Find \(i_\) (Current through 2kΩ):** The top branch has a total of **3kΩ**. The current in the top branch (\(i_{top}\)) is: \[ i_{top} = \frac{V}{3k\Omega} = \frac{38\,V}{3k\Omega} = .01267\,A = 12.7\,mA \] This is the current through the **entire top branch**, which splits through the 1kΩ and 2kΩ resistors (series). So, \[ i_ = i_{top} = 12.7\,mA \] --- ## **Step 3: Find \( i_g \) (Part B)** \( i_g \) is the total current supplied by the source: \[ i_g = i_{top} + i_{bottom} \] We already have \( i_{top} = 12.7\,mA \). For the bottom branch (3kΩ and 4kΩ in series = 7kΩ): \[ i_{bottom} = \frac{V}{7k\Omega} = \frac{38\,V}{7k\Omega} = .00543\,A = 5.43\,mA \] So, \[ i_g = 12.7\,mA + 5.43\,mA = 18.13\,mA \] --- ## **Step 4: Find Power Delivered (Part C)** Power delivered by the current source: \[ P = V \times i_g = 38\,V \times 18.13\,mA = .689\,W = 689\,mW \] --- ## **Summary Table** | Quantity | Value | Units | |----------|-------------------|--------| | \( i_ \) | 12.7 | mA | | \( i_g \) | 18.1 | mA | | \( P \) | 689 | mW | --- ### **Let me know if you need any step explained in more detail!**