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 $R_3 = 6$ $\Omega$.

Applied MathematicsCircuit AnalysisSuperposition TheoremResistorsCurrent Divider RuleOhm's Law
2025/7/2

1. Problem Description

The problem asks to find the current through the resistor R2R_2 (which has a resistance of 6 Ω\Omega) in the given circuit using the superposition theorem. The circuit contains two voltage sources E1=30E_1 = 30 V and E2=60E_2 = 60 V, and three resistors R1=12R_1 = 12 Ω\Omega, R2=6R_2 = 6 Ω\Omega, and R3=6R_3 = 6 Ω\Omega.

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 E1=30E_1 = 30 V and short circuit the voltage source E2E_2.
The circuit now consists of E1E_1, R1R_1, R2R_2, and R3R_3. R2R_2 and R3R_3 are in parallel. Let's find the equivalent resistance of R2R_2 and R3R_3:
R23=R2R3R2+R3=666+6=3612=3R_{23} = \frac{R_2 \cdot R_3}{R_2 + R_3} = \frac{6 \cdot 6}{6 + 6} = \frac{36}{12} = 3 Ω\Omega.
Now, R1R_1 and R23R_{23} are in series. The total resistance seen by E1E_1 is:
RT1=R1+R23=12+3=15R_{T1} = R_1 + R_{23} = 12 + 3 = 15 Ω\Omega.
The current flowing from E1E_1 is:
IT1=E1RT1=3015=2I_{T1} = \frac{E_1}{R_{T1}} = \frac{30}{15} = 2 A.
This current IT1I_{T1} flows through R1R_1. When it reaches the parallel combination of R2R_2 and R3R_3, it divides. Let I21I_{21} be the current flowing through R2R_2 due to E1E_1. Using the current divider rule:
I21=IT1R3R2+R3=266+6=2612=212=1I_{21} = I_{T1} \cdot \frac{R_3}{R_2 + R_3} = 2 \cdot \frac{6}{6 + 6} = 2 \cdot \frac{6}{12} = 2 \cdot \frac{1}{2} = 1 A.
Step 2: Consider the voltage source E2=60E_2 = 60 V and short circuit the voltage source E1E_1.
The circuit now consists of E2E_2, R1R_1, R2R_2, and R3R_3. R1R_1 and R2R_2 are in parallel. Let's find the equivalent resistance of R1R_1 and R2R_2:
R12=R1R2R1+R2=12612+6=7218=4R_{12} = \frac{R_1 \cdot R_2}{R_1 + R_2} = \frac{12 \cdot 6}{12 + 6} = \frac{72}{18} = 4 Ω\Omega.
Now, R3R_3 and R12R_{12} are in series. The total resistance seen by E2E_2 is:
RT2=R3+R12=6+4=10R_{T2} = R_3 + R_{12} = 6 + 4 = 10 Ω\Omega.
The current flowing from E2E_2 is:
IT2=E2RT2=6010=6I_{T2} = \frac{E_2}{R_{T2}} = \frac{60}{10} = 6 A.
This current IT2I_{T2} flows through R3R_3. When it reaches the parallel combination of R1R_1 and R2R_2, it divides. Let I22I_{22} be the current flowing through R2R_2 due to E2E_2. Using the current divider rule:
I22=IT2R1R1+R2=61212+6=61218=623=4I_{22} = I_{T2} \cdot \frac{R_1}{R_1 + R_2} = 6 \cdot \frac{12}{12 + 6} = 6 \cdot \frac{12}{18} = 6 \cdot \frac{2}{3} = 4 A.
Step 3: Find the total current I2I_2 through R2R_2.
I2=I21+I22=1+4=5I_2 = I_{21} + I_{22} = 1 + 4 = 5 A.

3. Final Answer

The current through the resistor R2R_2 is 5 A.

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