The image you provided shows a DC circuit with two resistors, $R_1 = 569 \, \Omega$ and $R_2 = 287 \, \Omega$, connected in series. The circuit has a voltage source of $74.31 \, V$. We are tasked with calculating the voltage, current, and power for each resistor.

1. Total Resistance

Since the resistors are in series, the total resistance $R_{\text{total}}$ is the sum of $R_1$ and $R_2$: $R_{\text{total}} = R_1 + R_2 = 569 \, \Omega + 287 \, \Omega = 856 \, \Omega$

2. Current in the Circuit

Using Ohm's law, the current $I$ through the circuit can be found by dividing the total voltage by the total resistance: $I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{74.31 \, V}{856 \, \Omega} \approx 0.0868 \, A = 86.8 \, mA$

3. Voltage Across Each Resistor

The voltage drop across each resistor can be found using Ohm's law ($V = IR$):

• For $R_1$: $V_1 = I \times R_1 = 0.0868 \, A \times 569 \, \Omega \approx 49.42 \, V$

• For $R_2$: $V_2 = I \times R_2 = 0.0868 \, A \times 287 \, \Omega \approx 24.89 \, V$

4. Power Dissipated by Each Resistor

The power dissipated by each resistor is given by $P = I^2 \times R$:

• For $R_1$: $P_1 = (0.0868 \, A)^2 \times 569 \, \Omega \approx 4.27 \, W$

• For $R_2$: $P_2 = (0.0868 \, A)^2 \times 287 \, \Omega \approx 2.16 \, W$

Summary of Results:

• $I = 86.8 \, mA$
• Voltage across $R_1 = 49.42 \, V$
• Voltage across $R_2 = 24.89 \, V$
• Power dissipated by $R_1 = 4.27 \, W$
• Power dissipated by $R_2 = 2.16 \, W$

Would you like further details on any of these steps or calculations?

Here are five related questions you might explore next:

1. What is the total power dissipated in the circuit?
2. How would the current change if $R_1$ were doubled?
3. How would the voltage across $R_2$ change if the total voltage was reduced to $50 \, V$?
4. What would happen to the circuit if $R_2$ was removed (open circuit)?
5. How does the power dissipation change if the resistors were arranged in parallel instead of series?

Tip: In a series circuit, the current remains the same through all components, but the voltage divides based on the resistance of each component.