Current Electricity - Kirchhoff’s Rules

Consider a simple closed loop containing a battery of $E = 12V$ and two resistors $R_1 = 2\Omega$ and $R_2 = 4\Omega$ connected in series. Find the current $I$ flowing through the circuit using Kirchhoff's Loop Rule.

Let the current in the loop be $I$. Starting from the negative terminal of the battery and traversing clockwise: $12 - I(2) - I(4) = 0$ $12 - 6I = 0$ $6I = 12$ $I = \frac{12}{6} = 2A$

According to the Loop Rule, the sum of the EMFs and potential drops must be zero. Since we move from the negative to the positive terminal of the cell, $E$ is $+12V$. Since we move in the direction of the current through the resistors, the potential changes are $-I R_1$ and $-I R_2$.

At a junction in an electrical circuit, four wires meet. Current $I_1 = 5A$ flows towards the junction, $I_2 = 2A$ flows away, and $I_3 = 4A$ flows towards the junction. Find the magnitude and direction of the current $I_4$ in the fourth wire.

Applying Kirchhoff's Junction Rule: $\sum I_{in} = \sum I_{out}$. Let $I_4$ be directed away from the junction. $I_1 + I_3 = I_2 + I_4$ $5 + 4 = 2 + I_4$ $9 = 2 + I_4$ $I_4 = 7A$

By the conservation of charge, the total current entering ($5A + 4A = 9A$) must equal the total current leaving ($2A + I_4$). Solving for $I_4$ gives $7A$ flowing away from the junction.

In a bridge circuit segment, three currents meet at junction $B$. If $I_{AB} = 4A$ (entering $B$) and $I_{BC} = 1.5A$ (leaving $B$ towards $C$), find the current $I_{BD}$ leaving junction $B$ towards $D$.

Applying Kirchhoff's Junction Rule at point $B$: $\sum I_{in} = \sum I_{out}$ $I_{AB} = I_{BC} + I_{BD}$ $4 = 1.5 + I_{BD}$ $I_{BD} = 4 - 1.5$ $I_{BD} = 2.5A$ The current flowing through branch $BD$ is $2.5A$.

According to the Junction Rule, charge cannot accumulate at a point. Therefore, the total current entering junction $B$ must be distributed among the outgoing branches.

In the given two-mesh circuit, find the currents $I_1$, $I_2$, and $I_3$ using Kirchhoff's rules. The circuit consists of two batteries $E_1 = 10V$ and $E_2 = 4V$, and three resistors $R_1 = 2\Omega$, $R_2 = 3\Omega$, and $R_3 = 5\Omega$. Current $I_1$ flows from $E_1$ through $R_1$, $I_2$ flows from $E_2$ through $R_2$, and $I_3$ is the sum flowing through $R_3$.

Applying Kirchhoff's First Rule at the junction: $I_3 = I_1 + I_2$

Applying Kirchhoff's Second Rule to Loop 1 (clockwise): $E_1 - I_1 R_1 - I_3 R_3 = 0$ $10 - 2I_1 - 5(I_1 + I_2) = 0$ $7I_1 + 5I_2 = 10 \quad \text{---(Eq. 1)}$

Applying Kirchhoff's Second Rule to Loop 2 (counter-clockwise): $E_2 - I_2 R_2 - I_3 R_3 = 0$ $4 - 3I_2 - 5(I_1 + I_2) = 0$ $5I_1 + 8I_2 = 4 \quad \text{---(Eq. 2)}$

Solving the simultaneous equations: Multiply Eq. 1 by 8 and Eq. 2 by 5: $56I_1 + 40I_2 = 80$ $25I_1 + 40I_2 = 20$

Subtracting the equations: $31I_1 = 60 \Rightarrow I_1 = \frac{60}{31} \approx 1.94A$

Substituting $I_1$ into Eq. 1: $7(1.94) + 5I_2 = 10$ $13.58 + 5I_2 = 10 \Rightarrow 5I_2 = -3.58 \Rightarrow I_2 = -0.716A$

Finally: $I_3 = I_1 + I_2 = 1.94 - 0.716 = 1.224A$

We use the Junction Rule to relate the three branch currents. Then, we apply the Loop Rule to the two independent loops. The negative sign for $I_2$ indicates that the actual current direction is opposite to the one assumed in the diagram.