Question

State Kirchhoff's rules. Using these rules, find the current flowing through branch FC in the given circuit.

Hint:

When applying Kirchhoff's Voltage Law (KVL), a powerful shortcut is the Nodal Potential Method. By setting one node (such as $F$) to $0\text{ V}$, you can find all other node potentials quickly without solving messy multi-variable loop equations.

Answer 1

Step 1: State Kirchhoff's rules:
Junction rule: the algebraic sum of the currents at any junction is zero, \(\sum I = 0\).
Loop rule: the algebraic sum of the emfs and potential drops around any closed loop is zero.

Step 2: Name the currents and the cell polarity:
Wires join A, F and E into one node. Wires join C and D into the second node. Three branches connect the two nodes.
Let \(I_1\) flow from A to B to C, \(I_2\) flow from F to C and \(I_3\) flow from E to D.
The longer plate of a cell is positive. So the 5 V cell has its positive plate toward A, the 3 V cell toward B and the 2 V cell toward F.
Junction rule at the left node: \(I_1 + I_2 + I_3 = 0\). So \(I_3 = -(I_1 + I_2)\).

Step 3: Loop ABCFA:
Go from A to B to C, then back from C to F to A.
The 5 V cell is crossed from positive to negative, giving \(-5\). The 3 V cell is crossed from negative to positive, giving \(+3\).
The resistors in the top branch give \(-1\cdot I_1 - 3 I_1\).
On the way back through the middle branch, the 1 ohm resistor is crossed against \(I_2\), giving \(+I_2\). The 2 V cell is crossed from negative to positive, giving \(+2\).
\[ -5 + 3 + 2 - 4I_1 + I_2 = 0 \;\Rightarrow\; I_2 = 4I_1 \]

Step 4: Loop FCDEF:
Go from F to C through the middle branch, then C to D, D to E and back to F.
The 2 V cell is crossed from positive to negative, giving \(-2\). The 1 ohm resistor gives \(-I_2\).
The 4 ohm resistor is crossed from D to E against \(I_3\), giving \(+4I_3\).
\[ -2 - I_2 + 4I_3 = 0 \]
Put \(I_3 = -(I_1 + I_2) = -5I_1\) and \(I_2 = 4I_1\):
\[ -2 - 4I_1 - 20I_1 = 0 \;\Rightarrow\; I_1 = -\frac{1}{12}\ \text{A} \]

Step 5: Find the current in FC:
\[ I_2 = 4I_1 = -\frac{1}{3}\ \text{A} \]
The negative sign shows that the current in FC actually flows from C to F.
Check: \(I_3 = -5I_1 = \dfrac{5}{12}\) A and \(I_1 + I_2 + I_3 = -\dfrac{1}{12} - \dfrac{4}{12} + \dfrac{5}{12} = 0\).

Final Answer:
The current in branch FC is \(\dfrac{1}{3}\ \text{A} \approx 0.33\ \text{A}\), flowing from C to F.