Find the currents I1 and I2 in the circuit in Fig

To find the currents \( I_1 \) and \( I_2 \) and the voltage \( V_0 \) in the circuit from Figure P2.19, we can apply Kirchhoff's laws and Ohm's law.

### Step 1: Identify the Resistances and Total Current

1. The circuit consists of:
- A current source of 1.5 A.
- A 150 Ω resistor.
- A 100 Ω resistor.
- A 250 Ω resistor.

### Step 2: Apply Kirchhoff's Current Law (KCL)

At the node where \( I_1 \) and \( I_2 \) meet:
\[
I = I_1 + I_2
\]
where \( I = 1.5 \, \text{A} \).

### Step 3: Analyze the Voltage Drops

Using Ohm's Law \( V = IR \), we can express the voltage across each resistor:

- For the 100 Ω resistor:
\[
V_1 = 100 \, \Omega \times I_1
\]

- For the 250 Ω resistor:
\[
V_2 = 250 \, \Omega \times I_2
\]

### Step 4: Apply Kirchhoff's Voltage Law (KVL)

Setting up KVL around the loop gives us:
\[
1.5 \, \text{A} \times 150 \, \Omega - V_1 - V_2 = 0
\]
Substituting for \( V_1 \) and \( V_2 \):
\[
150 \times 1.5 - 100 I_1 - 250 I_2 = 0
\]
\[
225 - 100 I_1 - 250 I_2 = 0 \quad (1)
\]

### Step 5: Substitute for \( I_1 \)

From KCL:
\[
I_1 = 1.5 - I_2 \quad (2)
\]

### Step 6: Solve the Equations

Substituting equation (2) into equation (1):

\[
225 - 100(1.5 - I_2) - 250 I_2 = 0
\]
\[
225 - 150 + 100 I_2 - 250 I_2 = 0
\]
\[
75 - 150 I_2 = 0
\]
\[
150 I_2 = 75
\]
\[
I_2 = 0.5 \, \text{A}
\]

Now substitute \( I_2 \) back into equation (2):
\[
I_1 = 1.5 - 0.5 = 1.0 \, \text{A}
\]

### Summary of Currents
- \( I_1 = 1.0 \, \text{A} \)
- \( I_2 = 0.5 \, \text{A} \)

### Step 7: Find the Voltage \( V_0 \)

To find the voltage \( V_0 \):
Using the voltage across the 150 Ω resistor:
\[
V_0 = 150 \times I_1 = 150 \times 1.0 = 150 \, \text{V}
\]

### Summary of Results
- \( I_1 = 1.0 \, \text{A} \)
- \( I_2 = 0.5 \, \text{A} \)
- \( V_0 = 150 \, \text{V} \)