Question

(a) The two Kirchhoff's rules are based on law of conservation of two physical quantities. Name the quantities.

(b) Find magnitude and direction of current in 1 \(\Omega\) resistor in the given circuit.

Hint:

The junction rule is conservation of charge and the loop rule is conservation of energy. Apply both rules to the two loops and the junction to get the current in 1 ohm.

Answer 1

Step 1: Kirchhoff's rules and conservation laws (part a):
The junction rule says the sum of currents entering a junction equals the sum leaving it. It is based on conservation of electric charge.
The loop rule says the sum of potential changes around a closed loop is zero. It is based on conservation of energy.

Step 2: Set up the currents (part b):
In the figure the 6 V cell has its long (positive) plate on the left, and the 9 V cell has its long plate on top.
Let \(I_2\) flow from P to Q through the left branch (cell 6 V and 3 \(\Omega\)). Let \(I_3\) flow from Q to P through the right branch (3 \(\Omega\), 9 V cell, 2 \(\Omega\)). Let \(I\) flow from P to Q through the 1 \(\Omega\) resistor.
Junction rule at P: \[ I_3=I_2+I \]

Step 3: Loop rule on the left loop:
Go from P through the left branch to Q and then back up through the 1 \(\Omega\) resistor. The cell raises the potential by 6 V, the 3 \(\Omega\) resistor drops it by \(3I_2\), and going up against \(I\) in the 1 \(\Omega\) resistor raises it by \(I\).
\[ 6-3I_2+I=0\ \Rightarrow\ I_2=\frac{6+I}{3} \]

Step 4: Loop rule on the right loop:
Go from Q through the right branch to P and then down through the 1 \(\Omega\) resistor.
\[ -3I_3+9-2I_3-I=0\ \Rightarrow\ I_3=\frac{9-I}{5} \]

Step 5: Solve:
Put \(I_2\) and \(I_3\) in the junction equation: \[ \frac{9-I}{5}=\frac{6+I}{3}+I \]
Multiply by 15: \(27-3I=30+5I+15I\), so \(23I=-3\).
\[ I=-\frac{3}{23}\ \text{A}\approx-0.13\ \text{A} \]
The negative sign means that the current is opposite to the direction we assumed.

Final Answer:
(a) Charge and energy. (b) The current in the 1 \(\Omega\) resistor is \(3/23\) A (about 0.13 A), flowing from Q to P. \[ \boxed{I=\frac{3}{23}\ \text{A}\approx0.13\ \text{A},\ \text{from Q to P}} \]