To solve this problem, let's go through each part step-by-step.

Given Data:

• Voltage source: $V = 9 \, \text{V}$
• Resistors:
  • $R_1 = 20 \, \Omega$
  • $R_2 = 20 \, \Omega$
  • $R_3 = 28 \, \Omega$
  • $R_4 = 7 \, \Omega$

(a) Redraw and Identify Nodes

The circuit can be redrawn by identifying each node in the circuit. Here, we have:

• Node 1: Top of $R_1$ and $R_2$, connected to the positive terminal of the battery.
• Node 2: Between $R_1$ and $R_3$.
• Node 3: Between $R_2$ and $R_4$.
• Node 4: Common node at the bottom, connecting the negative terminal of the battery and the bottom of $R_3$ and $R_4$.

(b) Find the Current Through $R_1$

To find the current through $R_1$, we first need to determine the equivalent resistance of the circuit.

1. Calculate Parallel Resistors:

   • $R_3$ and $R_4$ are in parallel. $\frac{1}{R_{34}} = \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{28} + \frac{1}{7} = \frac{1 + 4}{28} = \frac{5}{28}$ $R_{34} = \frac{28}{5} = 5.6 \, \Omega$

2. Calculate Series Combination:

   • Now, $R_1$, $R_2$, and $R_{34}$ are in series: $R_{\text{total}} = R_1 + R_2 + R_{34} = 20 + 20 + 5.6 = 45.6 \, \Omega$

3. Calculate Total Current:

   • Using Ohm's law: $I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{9 \, \text{V}}{45.6 \, \Omega} = 0.197 \, \text{A} \approx 0.2 \, \text{A}$

Since $R_1$, $R_2$, and $R_{34}$ are in series, the current through $R_1$ is the same as the total current. $I_{R_1} = 0.2 \, \text{A}$

(c) Find the Voltage Across $R_3$

To find the voltage across $R_3$, first determine the voltage across the $R_{34}$ parallel combination.

1. Voltage Across $R_{34}$:

   • Since $R_{34}$ is in series with $R_1$ and $R_2$, the voltage across $R_{34}$ can be found as: $V_{R_{34}} = I_{\text{total}} \times R_{34} = 0.2 \, \text{A} \times 5.6 \, \Omega = 1.12 \, \text{V}$

2. Voltage Division in Parallel Branch:

   • The voltage across $R_3$ is the same as $V_{R_{34}}$: $V_{R_3} = V_{R_{34}} = 1.12 \, \text{V}$

(d) Find the Power Delivered to $R_1$

The power delivered to $R_1$ can be calculated using the formula $P = I^2 R$.

1. Power in $R_1$: $P_{R_1} = I_{R_1}^2 \times R_1 = (0.2 \, \text{A})^2 \times 20 \, \Omega = 0.04 \times 20 = 0.8 \, \text{W}$

Summary of Answers:

• (a) The nodes are identified as described.
• (b) The current through $R_1$ is $0.2 \, \text{A}$.
• (c) The voltage across $R_3$ is $1.12 \, \text{V}$.
• (d) The power delivered to $R_1$ is $0.8 \, \text{W}$.

Would you like a more detailed explanation for any part?

Follow-up Questions:

1. How would the current change if the voltage source increased?
2. What if $R_3$ was removed from the circuit—how would that impact the total current?
3. How can the power across each resistor be calculated in parallel circuits?
4. What is the impact of combining resistors in parallel vs. series on the equivalent resistance?
5. How does Ohm's law apply differently in series vs. parallel circuits?

Tip:

In complex circuits, simplify by reducing series and parallel resistor combinations to understand current and voltage distribution.