To solve for the current through the circuit, let’s analyze the information given.

Circuit Parameters:

1. Voltage sources:
   • Generator with 120 V and internal resistance of 0.25 Ω.
   • A 24 V battery with an internal resistance of 0.15 Ω.
2. Resistors:
   • $R_1 = 11 \, \Omega$
   • $R_2 = 5.7 \, \Omega$

Steps to Calculate the Total Current:

1. Calculate the Total Resistance:

   • The resistances are connected in series, so the total resistance $R_{\text{total}}$ is the sum of all individual resistances: $R_{\text{total}} = R_1 + R_2 + 0.25 + 0.15 = 11 + 5.7 + 0.25 + 0.15 = 17.1 \, \Omega$

2. Calculate the Total Voltage:

   • Since the two voltage sources are in series, we add their voltages while considering their polarity. Here, they are additive: $V_{\text{total}} = 120 + 24 = 144 \, \text{V}$

3. Calculate the Current $I$:

   • Using Ohm's law, $I = \frac{V_{\text{total}}}{R_{\text{total}}}$: $I = \frac{144}{17.1} \approx 8.42 \, \text{A}$

Final Answer:

The current through the circuit is approximately 8.42 A.

Would you like further details or have any other questions?

Here are some related questions for deeper understanding:

1. How would the current change if the 24 V battery was reversed in polarity?
2. What would be the impact of removing the internal resistances of the voltage sources?
3. How does adding another resistor in series affect the circuit current?
4. Can we calculate the power dissipated in $R_1$ with the current found?
5. What happens to the current if $R_1$ and $R_2$ are swapped?

Tip: In series circuits, all components share the same current, which simplifies calculations compared to parallel circuits.