[Q13] The figure shows R₁ = 200 , R₂ = 100 , and the ideal batteries (no internal resistance) have emfs of &₁ = 7.0 V, E₂ = 6.0 V, and E3 = 5.0 V. Find the current through the resistor R₂ by using Kirchoff's laws on loop A. V = emf (1-e²+/RC) V = erf(1-2.718-1/100n()) 11.ov = (1-0,99005086) 11.or=(0.00994914²) 11.0v) 0.00994914) a E2 R₁ E ܐ E₁ #190² R₂ Va-V₂=E₁ + ₂ + Ez X -4 = √₂ + ²3 =-10.99 [C=10.99€ C

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**Question 13**

The figure shows \( R_1 = 200 \, \Omega \), \( R_2 = 100 \, \Omega \), and the ideal batteries (no internal resistance) have emfs of \( \varepsilon_1 = 7.0 \, \text{V} \), \( \varepsilon_2 = 6.0 \, \text{V} \), and \( \varepsilon_3 = 5.0 \, \text{V} \). Find the current through the resistor \( R_2 \) by using Kirchhoff's laws on loop A.

**Circuit Diagram:**
- The diagram is a simple loop circuit.
- It consists of three batteries with emfs \( \varepsilon_1 \), \( \varepsilon_2 \), \( \varepsilon_3 \).
- Two resistors \( R_1 \) and \( R_2 \) are connected in the loop.
- Points A and B are marked in the circuit.

**Equations:**

- Using Kirchhoff's law:
  \[
  V_{ab} - V_b = \varepsilon_1 + \varepsilon_2 + \varepsilon_3
  \]

- Rearrange:
  \[
  V_{ab} - V_b - \varepsilon_1 = \varepsilon_2 + \varepsilon_3
  \]

- Substitute values:
  \[
  V_{ab} = 6.0 \, \text{V} + 5.0 \, \text{V}
  \]
  \[
  V_{ab} = 11.0 \, \text{V}
  \]

**Calculations:**

- Voltage equation: 
  \[
  V = \text{emf} (1 - e^{-t/100RC})
  \]

- Substitute known values:
  \[
  11.0 \, \text{V} = (1 - 0.99005086)
  \]
  \[
  11.0 \, \text{V} = (0.009994914)
  \]

- Current equation:
  \[
  \frac{i}{c} = \left( 11.0 \, \text{V} \right) (0.009994914)
  \]
  \[
  i = C =
Transcribed Image Text:**Question 13** The figure shows \( R_1 = 200 \, \Omega \), \( R_2 = 100 \, \Omega \), and the ideal batteries (no internal resistance) have emfs of \( \varepsilon_1 = 7.0 \, \text{V} \), \( \varepsilon_2 = 6.0 \, \text{V} \), and \( \varepsilon_3 = 5.0 \, \text{V} \). Find the current through the resistor \( R_2 \) by using Kirchhoff's laws on loop A. **Circuit Diagram:** - The diagram is a simple loop circuit. - It consists of three batteries with emfs \( \varepsilon_1 \), \( \varepsilon_2 \), \( \varepsilon_3 \). - Two resistors \( R_1 \) and \( R_2 \) are connected in the loop. - Points A and B are marked in the circuit. **Equations:** - Using Kirchhoff's law: \[ V_{ab} - V_b = \varepsilon_1 + \varepsilon_2 + \varepsilon_3 \] - Rearrange: \[ V_{ab} - V_b - \varepsilon_1 = \varepsilon_2 + \varepsilon_3 \] - Substitute values: \[ V_{ab} = 6.0 \, \text{V} + 5.0 \, \text{V} \] \[ V_{ab} = 11.0 \, \text{V} \] **Calculations:** - Voltage equation: \[ V = \text{emf} (1 - e^{-t/100RC}) \] - Substitute known values: \[ 11.0 \, \text{V} = (1 - 0.99005086) \] \[ 11.0 \, \text{V} = (0.009994914) \] - Current equation: \[ \frac{i}{c} = \left( 11.0 \, \text{V} \right) (0.009994914) \] \[ i = C =
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