https://file.aiquickdraw.com/mathbot/file/1731976120344ilsibbp1.jpg

Let’s break this problem down step by step according to the instructions and the circuit diagram provided.

### Given Information:
- \( R_3 = 360 \, \Omega \)
- \( R_4 = 360 \, \Omega \)
- \( R_5 = 470 \, \Omega \)
- Voltage source \( E_1 = 5 \, \text{V} \)

---

### 1. **Connect a jumper between points A and D**
   This effectively creates a parallel circuit across \( R_3 \), \( R_4 \), and \( R_5 \).

---

### 2. **Calculate the total resistance (\( R_T \))**

Since \( R_3 \), \( R_4 \), and \( R_5 \) are in parallel, the formula for the equivalent resistance is:

\[
\frac{1}{R_T} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}
\]

Substitute the given values:

\[
\frac{1}{R_T} = \frac{1}{360} + \frac{1}{360} + \frac{1}{470}
\]

\[
\frac{1}{R_T} = 0.00278 + 0.00278 + 0.00213 = 0.00769
\]

\[
R_T = \frac{1}{0.00769} \approx 130 \, \Omega
\]

---

### 3. **Calculate the total current (\( I_T \))**

The total current is calculated using Ohm’s law:

\[
I_T = \frac{E_1}{R_T}
\]

\[
I_T = \frac{5}{130} \approx 0.0385 \, \text{A} \, (38.5 \, \text{mA})
\]

---

### 4. **Calculate the currents through \( R_3 \), \( R_4 \), and \( R_5 \)**

For resistors in parallel, the current through each resistor is calculated individually using Ohm’s law:

\[
I_R = \frac{V}{R}
\]

- \( I_{R3} = \frac{5}{360} \approx 0.01389 \, \text{A} \, (13.89 \, \text{mA}) \)
- \( I_{R4} = \frac{5}{360} \approx 0.01389 \, \text{A} \, (13.89 \, \text{mA}) \)
- \( I_{R5} = \frac{5}{470} \approx 0.01064 \, \text{A} \, (10.64 \, \text{mA}) \)

Check total current:
\[
I_T = I_{R3} + I_{R4} + I_{R5}
\]
\[
I_T = 13.89 + 13.89 + 10.64 \approx 38.5 \, \text{mA}
\]

---

### 5. **Calculate the power for each resistor and the total power**

The power dissipated by each resistor is calculated using the formula:

\[
P = I^2 R
\]

- \( P_{R3} = (0.01389)^2 \cdot 360 \approx 0.0694 \, \text{W} \, (69.4 \, \text{mW}) \)
- \( P_{R4} = (0.01389)^2 \cdot 360 \approx 0.0694 \, \text{W} \, (69.4 \, \text{mW}) \)
- \( P_{R5} = (0.01064)^2 \cdot 470 \approx 0.0529 \, \text{W} \, (52.9 \, \text{mW}) \)

Total power:
\[
P_T = P_{R3} + P_{R4} + P_{R5}
\]
\[
P_T = 0.0694 + 0.0694 + 0.0529 \approx 0.1917 \, \text{W} \, (191.7 \, \text{mW})
\]

---

### Summary of Results:
1. \( R_T = 130 \, \Omega \)
2. \( I_T = 38.5 \, \text{mA} \)
3. Currents:
   - \( I_{R3} = 13.89 \, \text{mA} \)
   - \( I_{R4} = 13.89 \, \text{mA} \)
   - \( I_{R5} = 10.64 \, \text{mA} \)
4. Powers:
   - \( P_{R3} = 69.4 \, \text{mW} \)
   - \( P_{R4} = 69.4 \, \text{mW} \)
   - \( P_{R5} = 52.9 \, \text{mW} \)
   - \( P_T = 191.7 \, \text{mW} \)

Let me know if you want detailed calculations for any part!  

---

**Questions for Further Understanding:**
1. How does connecting points A and D change the circuit's behavior?
2. What happens if \( R_5 \) is removed from the circuit?
3. Can you explain why the total power is the sum of the powers of individual resistors?
4. How would doubling \( E_1 \) affect \( I_T \) and the power dissipation?
5. What methods can be used to simplify more complex parallel and series circuits?

**Tip:** Always double-check your total current and resistance when working with parallel circuits to ensure consistency with Ohm’s law!

Concerning Figure 4.1, calculate the total resistance, total current, individual currents through each resistor, and power dissipated across the resistors using Ohm's law.

Learn how to calculate total resistance, current, and power dissipation in a parallel circuit with resistors R3, R4, and R5. Detailed calculations include Ohm's law and power formulas with practical examples.

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