Let's work through each part of the problem systematically.

Given Circuit:

• Voltage source $V_s = 100 \, \text{V}$
• Resistors: $R_1 = 5 \, \Omega$, $R_2 = 10 \, \Omega$, $R_3 = 7 \, \Omega$, $R_4 = 3 \, \Omega$

Step 1: Find the Equivalent Resistance ($R_{eq}$)

1. Combine $R_3$ and $R_4$ (which are in series):

$R_{34} = R_3 + R_4 = 7 \, \Omega + 3 \, \Omega = 10 \, \Omega$

2. Combine $R_{34}$ with $R_2$ (which are in parallel):

$\frac{1}{R_{234}} = \frac{1}{R_2} + \frac{1}{R_{34}} = \frac{1}{10 \, \Omega} + \frac{1}{10 \, \Omega} = \frac{1}{5 \, \Omega}$

Thus:

$R_{234} = 5 \, \Omega$

3. Combine $R_{234}$ with $R_1$ (which are in series):

$R_{eq} = R_1 + R_{234} = 5 \, \Omega + 5 \, \Omega = 10 \, \Omega$

Equivalent Resistance $R_{eq} = 10 \, \Omega$

Step 2: Find the Source Current $I_s$

Using Ohm’s law:

$I_s = \frac{V_s}{R_{eq}} = \frac{100 \, \text{V}}{10 \, \Omega} = 10 \, \text{A}$

Source Current $I_s = 10 \, \text{A}$

Step 3: Find the Voltage across the $3 \, \Omega$ Resistor

• The current flowing through the $3 \, \Omega$ resistor is the same as the current flowing through the parallel combination of $R_2$ and $R_{34}$.

$V_{34} = I_{234} \times R_4$

Since $R_2$ and $R_{34}$ are in parallel and have equal resistance:

$I_{234} = \frac{I_s}{2} = \frac{10 \, \text{A}}{2} = 5 \, \text{A}$

Therefore, the voltage across the $3 \, \Omega$ resistor is:

$V_4 = I_{234} \times R_4 = 5 \, \text{A} \times 3 \, \Omega = 15 \, \text{V}$

Voltage across $3 \, \Omega$ resistor $V_4 = 15 \, \text{V}$

Step 4: Verify Kirchhoff's Current Law (KCL)

KCL states that the sum of currents entering a junction must equal the sum of currents leaving the junction.

• At the junction between $R_1$, $R_2$, and $R_{34}$:
  • Current entering: $I_s = 10 \, \text{A}$
  • Currents leaving:
    • Through $R_2$: $5 \, \text{A}$
    • Through $R_{34}$: $5 \, \text{A}$

$I_s = I_{R_2} + I_{R_{34}} = 5 \, \text{A} + 5 \, \text{A} = 10 \, \text{A}$

KCL is verified.

Step 5: Verify Kirchhoff's Voltage Law (KVL) on the Left Loop

KVL states that the sum of the voltage drops in a closed loop must equal the sum of the voltage rises.

• Consider the left loop including the source $V_s$, $R_1$, and $R_2$:

$V_s = I_s \times R_1 + V_{R_2} = 10 \, \text{A} \times 5 \, \Omega + 50 \, \text{V} = 50 \, \text{V} + 50 \, \text{V} = 100 \, \text{V}$

KVL is verified.

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Summary of Answers:

1. Equivalent Resistance: $R_{eq} = 10 \, \Omega$
2. Source Current: $I_s = 10 \, \text{A}$
3. Voltage across 3-ohm resistor: $V_{4} = 15 \, \text{V}$
4. KCL: Verified
5. KVL on Left Loop: Verified

Would you like more details on any of these calculations, or have any questions?

Here are five related questions for further exploration:

1. What happens to the equivalent resistance if the $7 \, \Omega$ resistor is removed?
2. How would the current distribution change if the $10 \, \Omega$ resistor was replaced with a $20 \, \Omega$ resistor?
3. What is the power dissipated by the $3 \, \Omega$ resistor?
4. How does the circuit behavior change if the voltage source is doubled?
5. What would be the impact on $I_s$ if all resistances were halved?

Tip: Always double-check your current and voltage calculations when applying KVL and KCL to ensure accuracy.