The problem asks to find the current through the resistor $R_2$ (which has a resistance of 6 $\Omega$) in the given circuit using the superposition theorem. The circuit contains two voltage sources $E_1 = 30$ V and $E_2 = 60$ V, and three resistors $R_1 = 12$ $\Omega$, $R_2 = 6$ $\Omega$, and $R_3 = 6$ $\Omega$.
2025/7/2
1. Problem Description
The problem asks to find the current through the resistor (which has a resistance of 6 ) in the given circuit using the superposition theorem. The circuit contains two voltage sources V and V, and three resistors , , and .
2. Solution Steps
To solve this problem using superposition, we need to consider the contribution of each voltage source separately.
Step 1: Consider the voltage source V and short circuit the voltage source .
The circuit now consists of , , , and . and are in parallel. Let's find the equivalent resistance of and :
.
Now, and are in series. The total resistance seen by is:
.
The current flowing from is:
A.
This current flows through . When it reaches the parallel combination of and , it divides. Let be the current flowing through due to . Using the current divider rule:
A.
Step 2: Consider the voltage source V and short circuit the voltage source .
The circuit now consists of , , , and . and are in parallel. Let's find the equivalent resistance of and :
.
Now, and are in series. The total resistance seen by is:
.
The current flowing from is:
A.
This current flows through . When it reaches the parallel combination of and , it divides. Let be the current flowing through due to . Using the current divider rule:
A.
Step 3: Find the total current through .
A.
3. Final Answer
The current through the resistor is 5 A.