Let's break down the circuit in the image step by step to find the equivalent resistance, the current, and the voltage across each resistor.

Step 1: Analyze the Circuit

The circuit consists of a combination of series and parallel resistors:

1. Series combination: $R_1 = 30 \, \Omega$ and $R_3 = 60 \, \Omega$.
2. Series combination: $R_2 = 120 \, \Omega$ and $R_4 = 40 \, \Omega$.
3. These two sets are in parallel.

Step 2: Calculate Series Resistances

• The total resistance of the series combination $R_1$ and $R_3$: $R_{13} = R_1 + R_3 = 30 \, \Omega + 60 \, \Omega = 90 \, \Omega$
• The total resistance of the series combination $R_2$ and $R_4$: $R_{24} = R_2 + R_4 = 120 \, \Omega + 40 \, \Omega = 160 \, \Omega$

Step 3: Calculate the Parallel Resistance

Now, $R_{13} = 90 \, \Omega$ and $R_{24} = 160 \, \Omega$ are in parallel. The equivalent resistance of two resistors in parallel is given by: $\frac{1}{R_{\text{eq}}} = \frac{1}{R_{13}} + \frac{1}{R_{24}}$ Substituting the values: $\frac{1}{R_{\text{eq}}} = \frac{1}{90} + \frac{1}{160}$ To compute this: $\frac{1}{R_{\text{eq}}} = 0.01111 + 0.00625 = 0.01736$ Thus, the equivalent resistance $R_{\text{eq}}$ is: $R_{\text{eq}} = \frac{1}{0.01736} \approx 57.6 \, \Omega$

Step 4: Calculate the Total Current

The total voltage $V_T = 120 \, \text{V}$, and the equivalent resistance is $57.6 \, \Omega$. Using Ohm's Law $I = \frac{V}{R}$, the total current is: $I_T = \frac{120}{57.6} \approx 2.08 \, \text{A}$

Step 5: Find the Voltage and Current Across Each Resistor

• Current through $R_{13}$ and $R_{24}$: Since $R_{13}$ and $R_{24}$ are in parallel, they share the total voltage $V_T = 120 \, \text{V}$. The current in each branch is given by $I = \frac{V}{R}$: $I_{13} = \frac{120}{90} \approx 1.33 \, \text{A}, \quad I_{24} = \frac{120}{160} = 0.75 \, \text{A}$

• Voltage and current across $R_1$ and $R_3$:

  • $I_1 = I_3 = I_{13} = 1.33 \, \text{A}$
  • Voltage across $R_1$: $V_1 = I_1 \times R_1 = 1.33 \times 30 \approx 40 \, \text{V}$
  • Voltage across $R_3$: $V_3 = I_3 \times R_3 = 1.33 \times 60 \approx 80 \, \text{V}$

• Voltage and current across $R_2$ and $R_4$:

  • $I_2 = I_4 = I_{24} = 0.75 \, \text{A}$
  • Voltage across $R_2$: $V_2 = I_2 \times R_2 = 0.75 \times 120 = 90 \, \text{V}$
  • Voltage across $R_4$: $V_4 = I_4 \times R_4 = 0.75 \times 40 = 30 \, \text{V}$

Final Results:

• Total equivalent resistance: $R_{\text{eq}} = 57.6 \, \Omega$
• Total current: $I_T = 2.08 \, \text{A}$
• Current in branch $R_1, R_3$: $I_{13} = 1.33 \, \text{A}$
• Current in branch $R_2, R_4$: $I_{24} = 0.75 \, \text{A}$
• Voltages:
  • $V_1 = 40 \, \text{V}, V_3 = 80 \, \text{V}, V_2 = 90 \, \text{V}, V_4 = 30 \, \text{V}$

Would you like further details on any of these calculations? Here are some related questions:

1. How would the total resistance change if one resistor value increased?
2. What happens to the total current if the voltage source is doubled?
3. How does the power dissipated in each resistor relate to the current and resistance?
4. Could you calculate the power consumed by the entire circuit?
5. What are the implications of connecting more resistors in parallel in this network?

Tip: When dealing with parallel and series circuits, always simplify step by step to avoid mistakes.