Question

Find the currents through the resistors $R_1, R_2$, and $R_3$ with the help of the given circuit. Internal resistances of the cells are negligible.

Hint:

If current comes out negative, it means the actual direction is opposite to the assumed direction in KVL/KCL equations.

Answer 1

Step 1: Identify given values.
\[ R_1 = 10 \, \Omega, \quad R_2 = 20 \, \Omega, \quad R_3 = 10 \, \Omega. \] Two batteries: $30 \, \text{V}$ (with $R_1$), $10 \, \text{V}$ (with $R_3$).
Step 2: Apply Kirchhoff’s current law (KCL).
At the common junction: \[ I_1 = I_2 + I_3. \]
Step 3: Apply Kirchhoff’s voltage law (KVL).
Loop 1 (with $R_1$ and $R_2$): \[ 30 - 10I_1 - 20I_2 = 0. \] \[ 10I_1 + 20I_2 = 30. \quad (1) \] Loop 2 (with $R_2$ and $R_3$): \[ 10 - 10I_3 - 20I_2 = 0. \] \[ 10I_3 + 20I_2 = 10. \quad (2) \]
Step 4: Use current relation.
$I_1 = I_2 + I_3$. Substituting into (1): \[ 10(I_2 + I_3) + 20I_2 = 30. \] \[ 10I_2 + 10I_3 + 20I_2 = 30. \] \[ 30I_2 + 10I_3 = 30. \quad (3) \]
Step 5: Solve equations (2) and (3).
From (2): \[ 10I_3 + 20I_2 = 10. \] From (3): \[ 30I_2 + 10I_3 = 30. \] Subtract equations: \[ (30I_2 + 10I_3) - (20I_2 + 10I_3) = 30 - 10. \] \[ 10I_2 = 20 \quad \Rightarrow \quad I_2 = 2 \, \text{A}. \] Substitute in (2): \[ 10I_3 + 20(2) = 10. \] \[ 10I_3 + 40 = 10 \quad \Rightarrow \quad I_3 = -3 \, \text{A}. \] Negative sign means actual direction is opposite to assumed. So, \[ I_1 = I_2 + I_3 = 2 + (-3) = -1 \, \text{A}. \] Thus, \[ I_1 = -1 \, \text{A}, \quad I_2 = 2 \, \text{A}, \quad I_3 = -3 \, \text{A}. \]
Step 6: Final Answer.
Hence, actual directions of $I_1$ and $I_3$ are opposite to assumed. Magnitudes: \[ I_1 = 1 \, \text{A}, \quad I_2 = 2 \, \text{A}, \quad I_3 = 3 \, \text{A}. \] ---