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The circuit given in the image shows two resistors \( R_1 \) and \( R_2 \), both connected in series to an ideal battery with an emf \( \varepsilon = 11.3 \, \text{V} \). The values provided are \( R_1 = 6.69 \, \Omega \) and \( R_2 = 20.1 \, \Omega \).

### (a) Calculation of Current \( i_1 \)

In a series circuit, the total resistance \( R_{\text{total}} \) is simply the sum of the individual resistances:

\[
R_{\text{total}} = R_1 + R_2
\]

Substituting the given values:

\[
R_{\text{total}} = 6.69 \, \Omega + 20.1 \, \Omega = 26.79 \, \Omega
\]

Now, apply Ohm's law to find the current \( i_1 \):

\[
i_1 = \frac{\varepsilon}{R_{\text{total}}}
\]

\[
i_1 = \frac{11.3 \, \text{V}}{26.79 \, \Omega} \approx 0.4219 \, \text{A}
\]

Thus, the magnitude of the current \( i_1 \) is approximately **0.4219 A**.

---

### (b) Energy Dissipated by All Resistors

The energy dissipated by a resistor is given by the formula:

\[
E = I^2 R t
\]

Where:
- \( I \) is the current through the circuit (same for both resistors in a series circuit),
- \( R \) is the resistance of each resistor,
- \( t \) is the time in seconds.

First, convert the time \( 0.965 \, \text{min} \) into seconds:

\[
t = 0.965 \times 60 = 57.9 \, \text{seconds}
\]

Now, calculate the energy dissipated by both resistors:

\[
E_{\text{total}} = I^2 \cdot R_1 \cdot t + I^2 \cdot R_2 \cdot t
\]

\[
E_{\text{total}} = (0.4219)^2 \cdot 6.69 \cdot 57.9 + (0.4219)^2 \cdot 20.1 \cdot 57.9
\]

Calculating each term:

\[
E_1 = (0.4219)^2 \cdot 6.69 \cdot 57.9 \approx 65.15 \, \text{J}
\]

\[
E_2 = (0.4219)^2 \cdot 20.1 \cdot 57.9 \approx 195.73 \, \text{J}
\]

The total energy dissipated by all four resistors is:

\[
E_{\text{total}} \approx 65.15 + 195.73 = 260.88 \, \text{J}
\]

Thus, the total energy dissipated is approximately **260.88 J**.

---

Would you like further clarification on any of these calculations?

Here are some related questions for further study:
1. What is the total power dissipated in the circuit?
2. How does changing \( R_1 \) affect the current?
3. What happens to the energy dissipation if the time is doubled?
4. What is the potential drop across each resistor?
5. How would the circuit change if the resistors were connected in parallel?

**Tip**: For series circuits, remember that the current is the same through all resistors, while the voltage divides according to the resistances.

In the figure R1 = 6.69 Ω, R2 = 20.1 Ω, and the ideal battery has emf ε = 11.3 V. (a) What is the magnitude of current i1? (b) How much energy is dissipated by all four resistors in 0.965 min?

Learn how to calculate the magnitude of current and energy dissipated in a series circuit with resistors R1 = 6.69 Ω and R2 = 20.1 Ω, given an ideal battery with emf ε = 11.3 V. The solution covers Ohm's Law, energy dissipation, and step-by-step calculations for high school students in Grades 10-12.

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