Given: - \( R_1 = 3.00~k\Omega \) - \( R_2 = 2.00~k\Omega \) - \( R_3 = 8.00~k\Omega \) - \( \epsilon_1 = 77.0~V \) - \( \epsilon_2 = 57.0~V \) - \( \epsilon_3 = 77.0~V \) Let's use Kirchhoff's rules: Let the currents in each branch be: - \( I_1 \) through \( R_1 \) (from a to b) - \( I_2 \) through \( R_2 \) (from f to c) - \( I_3 \) through \( R_3 \) (from c to d) **Junction Rule at point c:** \[ I_1 = I_2 + I_3 \] ### 1. Loop 1: \( a \to f \to c \to b \to a \) \[ -\epsilon_1 + I_1 R_1 + I_2 R_2 + \epsilon_2 = 0 \] \[ I_1 R_1 + I_2 R_2 = \epsilon_1 - \epsilon_2 \] \[ 3I_1 + 2I_2 = 77 - 57 = 20 \quad \text{(in kΩ and V)} \] \[ 3I_1 + 2I_2 = 20 \tag{1} \] ### 2. Loop 2: \( c \to d \to e \to f \to c \) \[ -\epsilon_2 + I_3 R_3 - \epsilon_3 - I_2 R_2 = 0 \] \[ I_3 R_3 - I_2 R_2 = \epsilon_2 + \epsilon_3 \] \[ 8I_3 - 2I_2 = 57 + 77 = 134 \] \[ 8I_3 - 2I_2 = 134 \tag{2} \] ### 3. Loop 3: \( a \to b \to c \to d \to e \to f \to a \) Alternatively, use the two equations above and the junction equation: \[ I_1 = I_2 + I_3 \implies I_2 = I_1 - I_3 \] ### Substitute \( I_2 = I_1 - I_3 \) into (1) and (2): From (1): \[ 3I_1 + 2(I_1 - I_3) = 20 \] \[ 3I_1 + 2I_1 - 2I_3 = 20 \] \[ 5I_1 - 2I_3 = 20 \tag{3} \] From (2): \[ 8I_3 - 2(I_1 - I_3) = 134 \] \[ 8I_3 - 2I_1 + 2I_3 = 134 \] \[ 10I_3 - 2I_1 = 134 \tag{4} \] ### Now solve equations (3) and (4): From (3): \( 5I_1 - 2I_3 = 20 \) From (4): \( -2I_1 + 10I_3 = 134 \) Multiply (3) by 5: \[ 25I_1 - 10I_3 = 100 \] Add (4): \[ (25I_1 - 10I_3) + (-2I_1 + 10I_3) = 100 + 134 \\ 23I_1 = 234 \\ I_1 = \frac{234}{23} = 10.174~\text{mA} \] Plug \( I_1 \) into (3): \[ 5(10.174) - 2I_3 = 20 \\ 50.87 - 2I_3 = 20 \\ -2I_3 = 20 - 50.87 = -30.87 \\ I_3 = \frac{30.87}{2} = 15.44~\text{mA} \] ### **Final Answer:** **The current in resistor \( R_3 \) is:** \[ \boxed{15.44~\text{mA}} \] **(to two decimal places, as required)** --- If you need the direction: - The result is positive, so the current flows from \( c \) to \( d \) through \( R_3 \) as per our assumed direction.