Electrical Circuit Analysis Practice

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Difficulty

Direct

1.1Differentiate a Charge Function

Exam I | Problem 1.1 |

$$ q(t) = 4t^2 + 3t $$

in coulombs, what is the current at $t = 2$ s?

Solution

Current is the derivative of charge:

i(t) = \frac{dq}{dt} = 8t + 3

Evaluate at $t = 2$:

$$ i(2) = 8(2) + 3 = 19 $$

So the current is $19$ A.

1.2Compute Power with Passive Sign Convention

Exam I | Problem 1.2 |

A device has $v = 9$ V across it, and the current of $2$ A enters the positive-labeled terminal.

What is the power absorbed by the device?

Solution

Under the passive sign convention,

$$ p = vi $$

p = 9 \cdot 2 = 18

The device absorbs $18$ W.

1.3Find Current from Ohm's Law

Exam I | Problem 1.3 |

Ohm's Law, Resistors

A $6~\Omega$ resistor has $24$ V across it.

What current flows through the resistor?

Solution

Use Ohm's law:

$$ v = Ri $$

Solve for current:

i = \frac{v}{R} = \frac{24}{6} = 4

So the current is $4$ A.

1.4Add Series Resistors

Exam I | Problem 1.4 |

Series Resistance

Find the equivalent resistance of three series resistors:

3~\Omega,\quad 5~\Omega,\quad 7~\Omega

Solution

For series resistors, add the values:

R_{eq} = 3 + 5 + 7 = 15

So the equivalent resistance is $15~\Omega$.

1.5Combine Two Parallel Resistors

Exam I | Problem 1.5 |

Parallel Resistance

Find the equivalent resistance of $4~\Omega$ and $12~\Omega$ in parallel.

Solution

For two parallel resistors,

R_{eq} = \frac{R_1R_2}{R_1 + R_2}

R_{eq} = \frac{4 \cdot 12}{4 + 12} = \frac{48}{16} = 3

The equivalent resistance is $3~\Omega$.

1.6Use a Voltage Divider

Exam I | Problem 1.6 |

A $12$ V source drives two series resistors, $2~k\Omega$ and $4~k\Omega$.

If the output is taken across the $4~k\Omega$ resistor, what is $V_{out}$?

Solution

Use the voltage divider formula:

V_{out} = V_{in}\frac{R_2}{R_1 + R_2}

Substitute the values:

V_{out} = 12\frac{4}{2 + 4} = 12\cdot\frac{2}{3} = 8

So $V_{out} = 8$ V.

1.7Use a Current Divider

Exam I | Problem 1.7 |

A total current of $6$ A enters two parallel resistors, $3~\Omega$ and $6~\Omega$.

What current flows through the $3~\Omega$ branch?

Solution

For two parallel resistors, the current in one branch is

I_1 = I_{in}\frac{R_2}{R_1 + R_2}

Here,

I_{3\Omega} = 6\frac{6}{3 + 6} = 6\cdot\frac{2}{3} = 4

So the $3~\Omega$ branch carries $4$ A.

1.8Energy in a Capacitor

Exam I | Problem 1.8 |

A capacitor has $C = 2~\mu\text{F}$ and voltage $v = 10$ V.

How much energy is stored in the capacitor?

Solution

Use the capacitor energy formula:

w_C = \frac{1}{2}Cv^2

Substitute the values:

w_C = \frac{1}{2}(2\times 10^{-6})(10^2)

w_C = 10^{-4}\ \text{J}

So the stored energy is $1.0\times 10^{-4}$ J.

1.9Write an Element Impedance

Exam I | Problem 1.9 |

At angular frequency $\omega = 1000$ rad/s, what is the impedance of a $2~\mu\text{F}$ capacitor?

Solution

For a capacitor,

Z_C = \frac{1}{j\omega C}

Substitute the values:

Z_C = \frac{1}{j(1000)(2\times 10^{-6})}

Z_C = \frac{1}{j0.002} = -j500

So the impedance is $-j500~\Omega$.

1.10Balance Currents at a Node

Exam I | Problem 1.10 |

At a node, $2$ A and $5$ A enter, and $3$ A leaves through one branch.

How much current must leave through the other branch?

Solution

By KCL, the total current entering equals the total current leaving:

$$ 2 + 5 = 3 + I $$

$$ 7 = 3 + I $$

$$ I = 4 $$

The other branch must carry $4$ A leaving the node.

Difficulty

Integrated

2.1Write a Branch Current from Node Voltages

Exam II | Problem 2.1 |

Node $a$ is at $12$ V and node $b$ is at $5$ V.

If a $7~\Omega$ resistor connects $a$ to $b$, what is the current from $a$ to $b$?

Solution

Use the resistor current form:

i_{ab} = \frac{V_a - V_b}{R}

i_{ab} = \frac{12 - 5}{7} = 1

The current from $a$ to $b$ is $1$ A.

2.2Solve a Simple Supernode

Exam II | Problem 2.2 |

A supernode contains two unknown node voltages $V_a$ and $V_b$.

Node $a$ connects to ground through $6~\Omega$, node $b$ connects to ground through $3~\Omega$, and a $9$ V source connects $b$ to $a$ with the positive terminal at $a$.

If a total of $3$ A enters the supernode from an external source, what is $V_a$?

Solution

Use KCL on the supernode:

\frac{V_a}{6} + \frac{V_b}{3} = 3

The voltage source gives the constraint:

$$ V_a - V_b = 9 $$

Substitute $V_a = V_b + 9$ into KCL:

\frac{V_b + 9}{6} + \frac{V_b}{3} = 3

Multiply by $6$:

$$ V_b + 9 + 2V_b = 18 $$

$$ 3V_b = 9 $$

$$ V_b = 3 $$

Then

$$ V_a = 3 + 9 = 12 $$

So $V_a = 12$ V.

2.3Solve Two Mesh Currents

Exam II | Problem 2.3 |

Two clockwise mesh currents $i_1$ and $i_2$ share a $1~\Omega$ resistor.

The left mesh has a $1~\Omega$ resistor and a $12$ V source. The right mesh has a $1~\Omega$ resistor and no source.

What are $i_1$ and $i_2$?

Solution

Write KVL for each mesh.

Left mesh:

$$ 1i_1 + 1(i_1 - i_2) = 12 $$

which simplifies to

$$ 2i_1 - i_2 = 12 $$

Right mesh:

$$ 1i_2 + 1(i_2 - i_1) = 0 $$

which simplifies to

$$ -i_1 + 2i_2 = 0 $$

From the second equation,

$$ i_1 = 2i_2 $$

Substitute into the first:

$$ 2(2i_2) - i_2 = 12 $$

$$ 3i_2 = 12 $$

$$ i_2 = 4 $$

Then

$$ i_1 = 8 $$

So $i_1 = 8$ A and $i_2 = 4$ A.

2.4Use a Supermesh

Exam II | Problem 2.4 |

Two clockwise mesh currents $i_1$ and $i_2$ share a branch with a $1$ A current source.

The current source forces

$$ i_2 - i_1 = 1 $$

The outer loop contains two $3~\Omega$ resistors and a $15$ V source.

What are $i_1$ and $i_2$?

Solution

Write KVL around the outer perimeter of the supermesh:

$$ 3i_1 + 3i_2 = 15 $$

The current-source constraint is

$$ i_2 - i_1 = 1 $$

From the constraint,

$$ i_2 = i_1 + 1 $$

Substitute into the KVL equation:

$$ 3i_1 + 3(i_1 + 1) = 15 $$

$$ 6i_1 + 3 = 15 $$

$$ 6i_1 = 12 $$

$$ i_1 = 2 $$

Then

$$ i_2 = 3 $$

So $i_1 = 2$ A and $i_2 = 3$ A.

2.5Apply Superposition to a Node

Exam II | Problem 2.5 |

A node is connected to ground through a $6~\Omega$ resistor.

The same node is also connected through another $6~\Omega$ resistor to a $12$ V source, and a $3$ A current source injects current into the node from ground.

What is the node voltage?

Solution

Use superposition.

First, keep the $12$ V source and open the current source. The node is then a divider between $12$ V and ground through two equal $6~\Omega$ resistors, so the node voltage is:

V_1 = 12\frac{6}{6+6} = 6

Next, keep the $3$ A source and short the $12$ V source. Then the node sees two $6~\Omega$ resistors in parallel:

R_{eq} = \frac{6\cdot 6}{6+6} = 3

So the voltage contribution is

V_2 = IR = 3 \cdot 3 = 9

Add the contributions:

$$ V = V_1 + V_2 = 6 + 9 = 15 $$

So the node voltage is $15$ V.

2.6Find a Thevenin Equivalent

Exam II | Problem 2.6 |

A $12$ V source feeds a $2~\Omega$ resistor in series with a node.

From that node to ground is a $4~\Omega$ resistor.

Find the Thevenin equivalent voltage and resistance seen at the node with respect to ground.

Solution

The open-circuit voltage is the divider voltage across the $4~\Omega$ resistor:

V_{th} = 12\frac{4}{2+4} = 8

To find $R_{th}$, deactivate the independent source. The $12$ V source becomes a short circuit, so the $2~\Omega$ and $4~\Omega$ resistors are both from the node to ground:

R_{th} = \frac{2\cdot 4}{2+4} = \frac{8}{6} = \frac{4}{3}

So the Thevenin equivalent is $V_{th} = 8$ V in series with $R_{th} = \frac{4}{3}~\Omega$.

2.7Find an RC Step Response

Exam II | Problem 2.7 |

A $12$ V source charges a $100~\mu\text{F}$ capacitor through a $3~k\Omega$ resistor.

The capacitor starts at $0$ V.

What is $v_C(t)$ at $t = RC$?

Solution

First find the time constant:

\tau = RC = (3000)(100\times 10^{-6}) = 0.3\ \text{s}

For a charging capacitor,

v_C(t) = v_C(\infty) + \bigl(v_C(0^+) - v_C(\infty)\bigr)e^{-t/\tau}

Here,

v_C(0^+) = 0,\qquad v_C(\infty) = 12

v_C(t) = 12\bigl(1 - e^{-t/0.3}\bigr)

At $t = \tau$,

v_C(\tau) = 12(1 - e^{-1})

which is about $7.6$ V.

2.8Find a Series AC Current

Exam II | Problem 2.8 |

A source of $10\angle 0^\circ$ V at $\omega = 1000$ rad/s drives a series $100~\Omega$ resistor and $0.1$ H inductor.

What is the current phasor?

Solution

Find each impedance:

$$ Z_R = 100 $$

Z_L = j\omega L = j(1000)(0.1) = j100

So the total impedance is

$$ Z = 100 + j100 $$

Apply phasor Ohm's law:

\tilde{I} = \frac{\tilde{V}}{Z} = \frac{10}{100+j100}

Multiply by the complex conjugate:

\tilde{I} = \frac{10(100-j100)}{100^2+100^2}

\tilde{I} = 0.05 - j0.05

In polar form, this is

\tilde{I} \approx 0.0707\angle -45^\circ\ \text{A}

Difficulty

Applied

3.1Transform a Source to Simplify a Load

Final | Problem 3.1 |

A $12$ V source in series with a $6~\Omega$ resistor drives a $6~\Omega$ load.

Use source transformation to find the current through the load.

Solution

Convert the voltage source and series resistor to a Norton equivalent:

I_N = \frac{V_s}{R} = \frac{12}{6} = 2

So the source becomes a $2$ A current source in parallel with $6~\Omega$.

The load is also $6~\Omega$, so the two parallel resistors split the current equally:

I_L = 2\frac{6}{6+6} = 1

So the load current is $1$ A.

3.2Compute Load Power from a Thevenin Model

Final | Problem 3.2 |

A sensor port is modeled by a Thevenin equivalent of $18$ V in series with $3~\Omega$.

If it drives a $6~\Omega$ load, what power does the load absorb?

Solution

The total series resistance is

R_{tot} = 3 + 6 = 9

So the load current is

I = \frac{18}{9} = 2

The load power is

P_L = I^2R_L = 2^2 \cdot 6 = 24

So the load absorbs $24$ W.

3.3Estimate an RL Current After a Switch Closes

Final | Problem 3.3 |

A $10$ V source, $5~\Omega$ resistor, and $0.5$ H inductor are connected in series when a switch closes at $t=0$.

The inductor current is initially zero.

What is $i_L(0.2\ \text{s})$?

Solution

First find the time constant:

\tau = \frac{L}{R} = \frac{0.5}{5} = 0.1\ \text{s}

The final current is

i_L(\infty) = \frac{10}{5} = 2

So the current is

i_L(t) = 2\bigl(1 - e^{-t/0.1}\bigr)

At $t = 0.2$ s,

i_L(0.2) = 2(1 - e^{-2})

which is about $1.73$ A.

3.4Find Real Power and Power Factor

Final | Problem 3.4 |

A $120$ V rms source supplies a load current of $4$ A rms that lags the voltage by $30^\circ$.

What are the real power and the power factor?

Solution

The power factor is

\text{pf} = \cos 30^\circ = 0.866

Real power is

P = VI\cos\theta = 120 \cdot 4 \cdot 0.866

P \approx 416\ \text{W}

So the load has real power about $416$ W and power factor $0.866$ lagging.

3.5Evaluate an Inverting Amplifier

Final | Problem 3.5 |

An ideal inverting op-amp has $R_{in} = 2~k\Omega$ and $R_f = 8~k\Omega$.

If $V_{in} = 0.5$ V, what is the output voltage?

Solution

For an ideal inverting amplifier,

V_{out} = -\frac{R_f}{R_{in}}V_{in}

Substitute the values:

V_{out} = -\frac{8}{2}(0.5)

V_{out} = -2

So the output is $-2$ V.

Difficulty

Challenge

4.1Maximize Power to a Load

Final | Problem 4.1 |

A linear network seen from two terminals has a Thevenin equivalent of $20$ V in series with $5~\Omega$.

What load resistance maximizes the power transfer, and what is the maximum load power?

Solution

Maximum power transfer occurs when

R_L = R_{th}

So the best load is

R_L = 5~\Omega

The maximum power is

P_{max} = \frac{V_{th}^2}{4R_{th}} = \frac{20^2}{4\cdot 5}

P_{max} = 20\ \text{W}

So $R_L = 5~\Omega$ and the maximum load power is $20$ W.

4.2Use Continuity in an RC Transient

Final | Problem 4.2 |

A capacitor of $200~\mu\text{F}$ is initially at $6$ V.

At $t = 0$, it is connected through a $1~k\Omega$ resistor to an $18$ V source.

Find $v_C(t)$ and the time when the capacitor reaches $12$ V.

Solution

The capacitor voltage cannot change instantly, so

$$ v_C(0^+) = 6 $$

The final value is

v_C(\infty) = 18

The time constant is

\tau = RC = (1000)(200\times 10^{-6}) = 0.2\ \text{s}

v_C(t) = 18 + (6 - 18)e^{-t/0.2}

v_C(t) = 18 - 12e^{-t/0.2}

To find when $v_C(t) = 12$ V:

12 = 18 - 12e^{-t/0.2}

12e^{-t/0.2} = 6

e^{-t/0.2} = \frac{1}{2}

t = 0.2\ln 2

So the capacitor reaches $12$ V after $0.2\ln 2 \approx 0.139$ s.

4.3Find the Resonant Frequency of a Series RLC Circuit

Final | Problem 4.3 |

A series RLC circuit has $L = 50$ mH and $C = 200~\mu\text{F}$.

At what resonant frequency does the reactive part cancel?

Solution

For series resonance,

\omega_0 = \frac{1}{\sqrt{LC}}

Substitute the values:

\omega_0 = \frac{1}{\sqrt{(0.05)(200\times 10^{-6})}}

\omega_0 = \frac{1}{\sqrt{10^{-5}}} \approx 316.2\ \text{rad/s}

Convert to hertz:

f_0 = \frac{\omega_0}{2\pi} \approx \frac{316.2}{2\pi} \approx 50.3\ \text{Hz}

So the resonant frequency is about $50.3$ Hz.

4.4Evaluate an Inverting Summing Amplifier

Final | Problem 4.4 |

An ideal inverting summing amplifier has feedback resistor $R_f = 12~k\Omega$.

Two inputs are applied through $R_1 = 3~k\Omega$ with $V_1 = 1$ V and $R_2 = 6~k\Omega$ with $V_2 = 2$ V.

What is the output voltage?

Solution

For an ideal inverting summing amplifier,

V_{out} = -R_f\left(\frac{V_1}{R_1} + \frac{V_2}{R_2}\right)

Substitute the values:

V_{out} = -12\left(\frac{1}{3} + \frac{2}{6}\right)

V_{out} = -12\left(\frac{1}{3} + \frac{1}{3}\right)

V_{out} = -12\left(\frac{2}{3}\right) = -8

So the output voltage is $-8$ V.