Let's solve the problem step by step.

1. Understanding the circuit:

• $V_T = 120 \, \text{V}$ is the total voltage applied across the entire circuit.
• There are four resistors:
  • $R_1 = 30 \, \Omega$
  • $R_2 = 120 \, \Omega$
  • $R_3 = 60 \, \Omega$
  • $R_4 = 40 \, \Omega$

The circuit is a combination of series and parallel connections.

2. Simplify the parallel combinations:

• Resistors $R_3$ and $R_4$ are in parallel: $\frac{1}{R_{34}} = \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{60} + \frac{1}{40}$ First, find the equivalent resistance of $R_3$ and $R_4$: $\frac{1}{R_{34}} = \frac{1}{60} + \frac{1}{40} = \frac{2}{120} + \frac{3}{120} = \frac{5}{120}$ $R_{34} = \frac{120}{5} = 24 \, \Omega$

3. Equivalent resistance of the entire network:

Now, $R_1$, $R_2$, and $R_{34}$ are in series. The total equivalent resistance $R_{\text{eq}}$ is: $R_{\text{eq}} = R_1 + R_2 + R_{34} = 30 \, \Omega + 120 \, \Omega + 24 \, \Omega = 174 \, \Omega$

4. Find the total current:

Using Ohm's Law, the total current $I_T$ is: $I_T = \frac{V_T}{R_{\text{eq}}} = \frac{120 \, \text{V}}{174 \, \Omega} \approx 0.6897 \, \text{A}$

5. Voltage across each resistor:

• For resistors in series, the current is the same through all resistors. Now, calculate the voltage drop across each resistor.

• Voltage across $R_1$: $V_1 = I_T \times R_1 = 0.6897 \, \text{A} \times 30 \, \Omega \approx 20.69 \, \text{V}$

• Voltage across $R_2$: $V_2 = I_T \times R_2 = 0.6897 \, \text{A} \times 120 \, \Omega \approx 82.76 \, \text{V}$

• Voltage across $R_{34}$: $V_{34} = I_T \times R_{34} = 0.6897 \, \text{A} \times 24 \, \Omega \approx 16.55 \, \text{V}$

6. Current and voltage in $R_3$ and $R_4$:

Since $R_3$ and $R_4$ are in parallel, they have the same voltage, which is $V_{34}$.

• Current through $R_3$: $I_3 = \frac{V_{34}}{R_3} = \frac{16.55 \, \text{V}}{60 \, \Omega} \approx 0.276 \, \text{A}$

• Current through $R_4$: $I_4 = \frac{V_{34}}{R_4} = \frac{16.55 \, \text{V}}{40 \, \Omega} \approx 0.414 \, \text{A}$

Thus, the solution is:

• Equivalent resistance of the circuit: $174 \, \Omega$
• Total current: $0.6897 \, \text{A}$
• Voltage and current across each resistor:
  • $R_1$: $V_1 = 20.69 \, \text{V}, I_1 = 0.6897 \, \text{A}$
  • $R_2$: $V_2 = 82.76 \, \text{V}, I_2 = 0.6897 \, \text{A}$
  • $R_3$: $V_3 = 16.55 \, \text{V}, I_3 = 0.276 \, \text{A}$
  • $R_4$: $V_4 = 16.55 \, \text{V}, I_4 = 0.414 \, \text{A}$

Let me know if you'd like further details or explanations!

Follow-up questions:

1. What changes in the circuit's current if $R_1$ increases?
2. How would adding another resistor in parallel with $R_3$ affect the total resistance?
3. What happens to the total current if the total voltage is reduced to 60 V?
4. How does the voltage distribution change if $R_2$ is replaced with a $60 \, \Omega$ resistor?
5. What would be the effect on power consumption if $R_4$ were replaced with a $20 \, \Omega$ resistor?

Tip: Remember, in parallel circuits, the voltage is the same across all components, but the current divides according to the resistance of each branch!