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!