# Problems based on Kirchhoff’s laws – Problems and solutions Kirchhoff postulated two basic laws called Kirchhoff’s laws. The two laws are Kirchhoff’s Current Law and Kirchhoff’s Voltage Law. The problems below are based on Kirchhoff’s laws.

## Problems based on Kirchhoff’s laws

Q1: From the figure below find the value Resistance “R1

Solution: First name all nodes, assign currents to all branches and ground the node as shown in figure below \\ \\ we have assumed \\Current I in branch “ab” \\ I1 in branch “be” \\ I2 in branch “bc” and \\Ground node “e” as shown in above figure.

Now Apply Kirchhoff’s Voltage Law to loop “fabef” starting from node “f” and moving in direction “fabef”\\ 100 – 60I – I1R1– 20 = 0\\ 100 – 20= 60I + I1R1 \\ 80= 60I + I1R1 \;\;\;\;\;\;\;\;(1)\\

Now Apply Kirchoff’s current Law (KCL) at node “b”\\I = I1+ I2 \\ I2 = 0 because of open circuit \\ thus \\ I = I1 \;\;\;\;\;\;\;\;(2)

The Voltage across Resistance R1 \\70 – 20 = 50 \\thus\\\;\\50 = I1 R1\;\;\;\;\;\;\;\;(3)\\ Putting the value of I1R1 from equation (3) into equation (1) \\thus \\\;\\ 80 = 60I +50 \\\;\\ 80 – 50 =60I\\\;\\30 = 60I\\\;\\I=\frac{30}{60}=\frac{1}{2}\;\;\;\;\;\;\;\;\;(4)\\\;\\

From equation (2)\\\;\\I_{1}=I=\frac{1}{2}\\\;\\ Thus equation (3) becomes\\\;\\50=\frac{R_{1}}{2}\\\;\\R_{1}=100

Q2: In the circuit shown below find the value of V1 \\\;\;\;\;\;\;\;\; Assign Names to each node. Assuming mesh current “I1“, “I2“, “I3“, and “I4” to loop “I”, “II”, “III”, and “IV” respectively, as shown in figure below\\\;\\\;\;\;\;\; \\Applying Kirchhoff’s Voltage law to loop “I” and move in the direction of arrow starting from node “a” .

-7-2(I_{1}-I_{2})+V_{1}=0\\\;\\ V_{1}=2I_{1}-2I_{2}+7\;\;\;\;\;\;\;\;\;(1)

Now Apply Kirchhoff’s Voltage law to Loop “II” and moving in the direction of arrow, starting from node “c” \\\;\\5-2(I_{2}-I_{1})-6=0\\\;\\ -1=2I_{2}-2I_{1} \\\;\\1=2I_{1}-2I_{2}\;\;\;\;\;\;\;(2)\\\;\\ Putting the value of \;2I_{1}-2I_{2}\; in equation (1) from equation (2) Thus \\\;\\V_{1}=1+7=8V

Q3: In the circuit shown in figure find the value of voltage Vab \\\;\;\;\;\;\;\;\; Assign names of each node and loops as shown in figure. \\\;\\\;\;\;\;\; \\Let us assume Vab is the voltage across 0.2I1 current source in branch “ab” , “I2” is current through 3Ω resistance, “I3” and “I4” are the currents through branch “ec” and “ac” respectively.

Applying Kirchhoff’s current Law (KCL) at node “d” \\\;\\8=i_{1}+2\\\;\\i_{1}=6

Apply Kirchhoff’s current law (KCL) at node “b” \\\;\\i_{1}=0.2i_{1}+0.3i_{1}+i_{2}\\\;\\ i_{2}=0.5i_{1} \\\;\\ Putting the value of “I1” in above equation \\\;\\i_{2}=3

Applying KCL at node “a” \\\;\\2+i_{4}+0.2i_{1}=0\\\;\\ putting the value “I1” in above equation \\\;\\i_{4}=-3.2\\\;\\

Now Applying Kirchhoff’s Voltage Law to loop “II” i.e. “abca” and move in dircetion “a” to “b” to “c” to “a” starting from node “a” \\\;\\V_{ab}-3i_{2}-2i_{4}=0\\\;\\ Putting the values of “I2” and “I4\\\;\\V_{ab}-3\left ( 3 \right )-2\left ( -3.2 \right )=0\\\;\\V_{ab}=2.6 V

Q4: Consider the circuit shown in figure below find the relationship between voltage V1 and V2 \\\;\;\;\;\;\;\;\; Assign names of each node as shown in figure below.\\\;\\\;\;\;\;\; \\\;\\ Applying KVL to branch “ab” \\\;\\V_{1}-6i_{a}-8-V_{2}=0\\\;\\V_{1}=6i_{a}+8+V_{2} 