Solved Problems on Kirchhoff's Voltage and Current Laws (KVL & KCL)

Solution-1

• KCL at first node $A$ : $9 = i_0 + i_1 \Rightarrow i_1 = 9-i_0$

• KCL at second node $B$ : $i_1 = \dfrac{i_0}{4}+i_2 \Rightarrow i_1 = \dfrac{i_0}{4}+\dfrac{v_0}{8}$

• substituting the first in second, we get $$\begin{aligned} 9-i_0 & = \dfrac{i_0}{4}+\dfrac{v_0}{8} \\ \Rightarrow 9-\dfrac{v_0}{8} & = \dfrac{5i_0}{4} \end{aligned}$$

Problem-2

• Obtain the current $I$ in the network?

Solution-2

• Applying KCL at Node-A $I_1 = I-2$

• Now, $V_R = 2I_1 = 2I-4$

• On substitution $$\begin{aligned} 2(2I-4) & = 5I+4\\ \Rightarrow I & = -12~\mathrm{A} \end{aligned}$$

Problem-3

• Obtain the current $i$ in the network?

Solution-3

• KVL in second loop $4V_{ab} = 4i \Rightarrow i = V_{ab}$

• $V_b = i$

• $V_{ab} = V_a - V_b = 2.5 - i = i$

• Hence, $i = 1.25~\mathrm{A}$

Problem-4

• Determine the following in the circuit

  1. $V_{R2}$

  2. $V_2$ if $V_{R1}=1$

Solution-4

• Again applying KVL $$v_x +v_2+3-v_{R1}=0 \Rightarrow v_2 = -v_x-3+v_{R1} = -8~\mathrm{V}$$

Problem-5

• The voltage source in the given circuit has a current of 1A flowing out of its positive terminal into resistor $R_1$ .Calculate the current labelled $i_2$.

Solution-5

$$\begin{aligned} 1 &= i_2 -3 +7 \\ \Rightarrow i_2& = -3~\mathrm{A} \end{aligned}$$

Problem-6

• Determine the value of the voltage $v$ ?

Solution-6

• Applying KCL

  $$\begin{aligned} 1+2&=\dfrac{v}{5}+5+\dfrac{v}{5} \\ -2 & = \dfrac{2v}{5}\\ v & = -5~\mathrm{V} \end{aligned}$$

Problem-7

• Find the current $i_x$ in the circuit shown?

Solution-7

• Applying KCL at three nodes $$\begin{aligned} 4\mathrm{mA}+3i_1+\dfrac{V}{5~k\Omega}+\dfrac{V}{20~k\Omega}&=0\\ i_1 & = 3i_1+\dfrac{V}{20~k\Omega}\\ 3i_1+\dfrac{V}{20~k\Omega} & = i_x \end{aligned}$$

• Solving the first two and then putting in third, we get $$i_x = 0.57~\mathrm{A}$$

Problem-8

• For the given circuit, use the KVL to find the branch voltages $V_1$ to $V_4$?

Solution-8

• Applying KVL in 4 loops $$\begin{aligned} 3-V_1+V_3&=0\\ V_1+V_2+2&=0\\ -4-V_3-V_4&=0\\ V_4-2-5&=0 \end{aligned}$$

• $V_4=7\mathrm{V}~V_3=-11\mathrm{V}~V_1=-8\mathrm{V}~V_2=6\mathrm{V}$

Problem-9

• Determine the value of $v_x$ labelled in the circuit?

Solution-9

• Voltage at first node $2.3-500mA\times 7.3=-1.35\mathrm{V}$

• Applying KCL at first node $0.5+1.35=I_2$

• Therefore $v_x = -1.35-1.85\times 2=-5.05\mathrm{V}$

Problem-10

• The current $I_x$ in the circuit given ?

Solution-10

• Applying KCL at node $$\dfrac{V-1}{100}+\dfrac{V}{100} = 10~\mathrm{mA}$$

• Then, $$I_x = \dfrac{V}{100}$$

• $I_x = 10~\mathrm{mA}$