DC Circuit Analysis
Kirchhoffs Current Law
Learning Objectives:

Explain how to apply Kirchhoff Current Law (KCL) to electric circuits.

Be able to write KCL at every node in the circuit.

Be able to solve the KCL equations, especially for simple circuits.
The algebraic sum of the currents entering a node must equal the sum of the currents exiting a node.
If there is a junction where \(I_{_{1}} , I_{_{2}} , I_{_{3}}\) are the currents entering and \(I_{_{4}} , I_{_{5}}\) are currents leaving, then
\[I_{_{1}} + I_{_{2}} + I_{_{3}}  I_{_{4}}  I_{_{5}} = 0\]
Solved Example: 19101
Kirchhoff's current law states that:
A. Net current flow at the junction is positive
B. Algebraic sum of the currents meeting at the junction is zero
C. No current can leave the junction without some current entering it.
D. Total sum of currents meeting at the junction is zero.
Kirchhoff's current law (1st Law) states that current flowing into a node (or a junction) must be equal to current flowing out of it. This is a consequence of charge conservation.
Correct Answer: B
Solved Example: 19102
Kirchhoff's current law is applicable to only:
A. Junction in a network
B. Closed loops in a network
C. Electric circuits
D. Electronic circuits
Kirchhoff's Laws describe current in a node and voltage around a loop.
Correct Answer: A
Solved Example: 19103
Two components are in series if they have: (DFCCIL Executive Electrical 2016)
A. One current path
B. Two current paths
C. The same voltage
D. Different voltages
Correct Answer: A
Solved Example: 19104
When two components are in parallel, the circuit has:
A. One current path
B. Two current paths
C. The same power loss
D. Different power losses
Correct Answer: B
Solved Example: 19105
As a general rule, how are electronic components connected when placed into circuit?
A. In series
B. In parallel as
C. A combination of series and parallel
D. As neither series, parallel, nor seriesparallel
Correct Answer: C
Solved Example: 19106
What determines whether connected resistors are in series, parallel, or seriesparallel?
A. Current flow
B. Power source
C. Resistance
D. Voltage source
Correct Answer: A
Solved Example: 19107
Where can the voltage divider formula be used in a seriesparallel circuit?
A. With a group of seriesconnected resistors
B. A group of parallelconnected resistors
C. With a group of seriesparallel connected resistors
D. With seriesopposing power supplies
Correct Answer: A
Solved Example: 19108
Which of the following is the more useful in solving seriesparallel circuit problems?
A. Ohms' law
B. Gilbert's law
C. Kirchoff's current and voltage laws
D. Lenz's law
Correct Answer: C
Kirchhoffs Voltage Law
Learning Objectives:

Explain how to apply Kirchhoff Voltage Law to electric circuits.

Be able to write KVL for every loop in the circuit.

Be able to solve the KVL equations, especially for simple circuits.
The voltages around a closed path in a circuit must sum to zero.
Kirchhoff’s Voltage Law (or Kirchhoff’s Loop Rule) is a result of the electrostatic field being conservative. It states that the total voltage around a closed loop must be zero. If this were not the case, then when we travel around a closed loop, the voltages would be indefinite. So \[\large \sum V = 0\]
Sign Conventions:

If a resistance is travelled in the same direction as of current then the voltage drop across that resistance is considered negative.

If a voltage source is travelled from negative to positive, that voltage value is considered positive.
Solved Example: 19201
According to Kirchhoff's voltage law, the algebraic sum of all IR drops and e.m.fs. in any closed loop of a network is always:
A. Negative
B. Positive
C. Determined by battery e.m.fs.
D. Zero
Correct Answer: D
Solved Example: 19202
A closed path made by several branches of the network is known as:
A. Branch
B. Loop
C. Circuit
D. Junction
Correct Answer: B
Solved Example: 19203
Kirchhoff's voltage law is related to:
A. Junction currents
B. Battery e.m.fs.
C. IR drops
D. Both (B) and (C)
Correct Answer: D
Solved Example: 19204
Find the current supplied by 7 V source.
A. 0.99 A
B. 1.06 A
C. 1.21 A
D. 1.39 A
We will consider three loops with loop currents as I$_1$, I$_2$ and I$_3$ as shown in the following figure.
Applying Kirchhoff's Voltage rule to loop with i$_1$,
\[7  3(i_1  i_2)  1(i_1  i_3) 5 = 0\] \[4i_1 + 3 i_2 + i_3 = 2\]Applying Kirchhoff's Voltage rule to loop with i$_2$,
\[5 i_2  3 (i_2  i_3)  3(i_2  i_1) = 0\] \[3i_1  11 i_2 + 3i_3 = 0\]Applying Kirchhoff's Voltage rule to loop with i$_3$,
\[5  1(i_3  i_1)  3(i_3  i_2)  6 i_3 = 0\] \[i_ 1 + 3 i_2  10i_3 = 5\]Solving these three equations using determinants,
\begin{align*} D &= \begin{vmatrix} 4 & 3 &1\\ 3&11 &3 \\ 1& 3 &10 \end{vmatrix}\\ &= 285 \end{align*} \begin{align*} D_{i_1} &= \begin{vmatrix} 2 & 3 &1\\ 0&11 &3 \\ 5& 3 &10 \end{vmatrix}\\ &= 302 \end{align*} \[i_1 = \dfrac{D_{i_1}}{D} = \dfrac{302}{285} = 1.06\ A\]Correct Answer: B
Resistances in Series
Learning Objectives:

Recall and use R = V/I

Calculate the combined resistance of resistors connected in series.

Perform calculations on voltage divider circuits;

Define power as VI and apply the formula to calculate power dissipation in series circuits.
Components connected in series are only connected to each other at one point.
The same current flows through both components. The sum of the voltages across each component in the circuit equals the power supply voltage.
\[R_{\mathrm{series}} = R_1 + R_2 + R_3 + ......................\]
Series Direct Current Circuit Rules:

The same current flows through each part of a series circuit.

Total Resistance of a series circuit is equal to the sum of the individual resistances.

The total voltage across a series circuit is equal to the sum of the individual voltage drops.

The voltage drop across a resistor in a series circuit is proportional to the size of the resistor.

The total power dissipated in a series circuit is equal to the sum of the individual power dissipation.
Solved Example: 19301
If length of a conductor is doubled by keeping volume constant, then what is its new resistance if initial were 4 $\Omega$ ?
A. 8 $\Omega$
B. 4 $\Omega$
C. 2 $\Omega$
D. 16 $\Omega$
Correct Answer: D
Solved Example: 19302
A cell of e.m.f 2 V and internal resistance 0.5 $\Omega$ is connected across a resistor R. The current that flows is same as that, when a cell of e.m.f 1.5 V and internal resistance 0.3 $\Omega$ is connected across the same resistor. Then:
A. R = 0.3 $\Omega$
B. R = 0.6 $\Omega$
C. R = 0.5 $\Omega$
D. R = 0.75 $\Omega$
\begin{align*} \dfrac{2}{0.5 + R} &= \dfrac{1.5}{0.3 + R}\\ R &= 0.3\ \Omega \end{align*}
Correct Answer: A
Resistances in Parallel
Learning Objectives:

Calculate the combined resistance of resistors connected in parallel.

Perform calculations on current divider circuits;

Define power as VI and apply the formula to calculate power dissipation in parallel circuits.
The components sit side by side and connect to more than one other component.
The voltage is the same across all parallel components.
\[\dfrac{1}{R_{\mathrm{parallel}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + ......................\]
The sum of the currents in each branch of the circuit equals the current leaving the power supply.
Parallel Direct Current Circuit Rules:

The same voltage exists across each branch of a parallel circuit and is equal to the source voltage.

The current through a branch of a parallel network is inversely proportional to the amount of resistance of the branch.

The total current of a parallel circuit is equal to the sum of the currents of the individual branches of the circuit.

The total resistance of a parallel circuit is equal to the reciprocal of the sum of the reciprocals of the individual resistances of the circuit.

The total power dissipated in a parallel circuit is equal to the sum of the individual power dissipation.
Solved Example: 19401
Effective resistance of parallel combination of two resistors is $\dfrac{6}{4} \Omega$. If one of resistance is broken, then the resultant resistance becomes 2 $\Omega$. Then other resistance is:
A. 4 $\Omega$
B. 3 $\Omega$
C. 6 $\Omega$
D. 5 $\Omega$
\[\dfrac{1}{2} + \dfrac{1}{R} = \dfrac{1}{\dfrac{6}{4}}, \quad \dfrac{1}{R} = \dfrac{1}{6}, \quad R = 6 \ \Omega\]
Correct Answer: C
Solved Example: 19402
The currents into a junction flow along two paths. One current is 4 A and the other is 3 A. The total current out of the junction is:
A. 1 A
B. 7 A
C. Unknown
D. The larger of the two
Correct Answer: B
Solved Example: 19403
When an additional resistor is connected across an existing parallel circuit, the total resistance:
A. Remains the same
B. Decreases by the value of the added resistor
C. Increases by the value of the added resistor
D. Decreases
Correct Answer: D
Solved Example: 19404
When a 1.6 $k\Omega$ resistor and a 120 $\Omega$ resistor are connected in parallel, the total resistance is:
A. Greater than 1.6 $k\Omega$
B. Greater than 120 $\Omega$ but less than 1.6 $k\Omega$
C. Less than 120$\Omega$ but greater than 100$\Omega$
D. Less than 100 $\Omega$
Correct Answer: C
Solved Example: 19405
The power dissipation in each of four parallel branches is 1.2 W. The total power dissipation is:
A. 1.2 W
B. 4.8 W
C. 0.3 W
D. 12 W
Power dissipated in parallel circuit overall is sum of power dissipated in individual branches.
Correct Answer: B
Solved Example: 19406
What is the total resistance, if a seriesparallel circuit using $R_1$ =30 $\Omega$ in series with a parallel combination of $R_2$ = 30 $\Omega$ and $R_3$ = 30 $\Omega$ ?
A. 10 $\Omega$
B. 20 $\Omega$
C. 45 $\Omega$
D. 90 $\Omega$
Correct Answer: C
Solved Example: 19407
Find the current I$_x$ using superposition theorem.
A. 0.96 A
B. 4.99 A
C. 5.21 A
D. 5.39 A
In Principle of Superposition, only one source is active at a time. Case I:
36 V is active, 12 V is short circuited, 6 A is open circuited.
5 $\Omega$ is parallel to 10 $\Omega$ = 3.33 $\Omega$
3.33 $\Omega$ is in series with 5 $\Omega$ = 8.33 $\Omega$
\[I_x = \dfrac{36}{8.33} = 4.32 A\] Case II:12 V is active, 36 V is short circuited, 6 A is open circuited.
5 $\Omega$ is parallel to 10 $\Omega$ = 3.33 $\Omega$
3.33 $\Omega$ is in series with 5 $\Omega$ = 8.33 $\Omega$
Total current = $\dfrac{12}{8.33}$ = 1.44 A
Current entering 5 $\Omega$ = $\dfrac{2}{3}$ $\times$ 1.44 A = 0.96 A
I$_x$ = 0.96 A
Case III:6 A is active, 12 V is short circuited, 36 V is short circuited.
5 $\Omega$ is parallel to 10 $\Omega$ = 3.33 $\Omega$
3.33 $\Omega$ is in parallel with 5 $\Omega$
Current entering 5 $\Omega$ = 6 $\times$ $\dfrac{3.33}{3.33 + 5}$
I$_x$ = 2.40 A
Total I$_x$ = 4.32  0.96  2.40 = 0.96 A (From left to Right)
Correct Answer: A