State Kirchhoff’s laws. Apply these laws to find the values of current flowing in the three branches of the given circuit.
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When using Kirchhoff's Laws, make sure to carefully assign current directions and carefully apply the loop and junction rules. Solving the system of equations will give the correct values for currents in each branch of the circuit.
The given circuit has the following elements:
- A 5 V battery in series with a 2Ω resistor.
- A 10 V battery in series with a 1Ω resistor and a 2Ω resistor.
Step 1: Assign currents to the branches:
Let the current in the branch with the 5V source and 2Ω resistor be \( I_1 \), the current in the branch with the 10V source and 1Ω resistor be \( I_2 \), and the current in the branch with the 2Ω resistor be \( I_3 \).
Step 2: Apply Kirchhoff's Voltage Law (KVL):
For loop 1 (containing the 5V source and 2Ω resistor):
\[
5 - 2I_1 = 0 \quad \Rightarrow \quad I_1 = \frac{5}{2} \, \text{A}
\]
For loop 2 (containing the 10V source, 1Ω resistor, and 2Ω resistor):
\[
10 - 1I_2 - 2I_3 = 0
\]
Apply Kirchhoff's Current Law (KCL):
At the junction where the currents meet, we have:
\[
I_1 = I_2 + I_3
\]
Substitute \( I_1 = \frac{5}{2} \) into the equation:
\[
\frac{5}{2} = I_2 + I_3
\]
Step 3: Solve the system of equations:
We have two equations:
1. \( 10 - I_2 - 2I_3 = 0 \)
2. \( \frac{5}{2} = I_2 + I_3 \)
From the first equation:
\[
I_2 = 10 - 2I_3
\]
Substitute this into the second equation:
\[
\frac{5}{2} = (10 - 2I_3) + I_3
\]
\[
\frac{5}{2} = 10 - I_3
\]
\[
I_3 = 10 - \frac{5}{2} = \frac{15}{2} \, \text{A}
\]
Now substitute \( I_3 = \frac{15}{2} \) back into \( I_2 = 10 - 2I_3 \):
\[
I_2 = 10 - 2 \times \frac{15}{2} = 10 - 15 = -5 \, \text{A}
\]
Thus, the current values are:
- \( I_1 = \frac{5}{2} \, \text{A} \)
- \( I_2 = -5 \, \text{A} \)
- \( I_3 = \frac{15}{2} \, \text{A} \)
The negative sign in \( I_2 \) indicates that the direction of current in the second branch is opposite to the assumed direction.
