Pomodoro
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Difficulty
If
in coulombs, what is the current at $t = 2$ s?
Solution
Current is the derivative of charge:
Evaluate at $t = 2$:
So the current is $19$ A.
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,
so
The device absorbs $18$ W.
A $6~\Omega$ resistor has $24$ V across it.
What current flows through the resistor?
Solution
Use Ohm's law:
Solve for current:
So the current is $4$ A.
Find the equivalent resistance of three series resistors:
Solution
For series resistors, add the values:
So the equivalent resistance is $15~\Omega$.
Find the equivalent resistance of $4~\Omega$ and $12~\Omega$ in parallel.
Solution
For two parallel resistors,
So
The equivalent resistance is $3~\Omega$.
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:
Substitute the values:
So $V_{out} = 8$ V.
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
Here,
So the $3~\Omega$ branch carries $4$ A.
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:
Substitute the values:
So the stored energy is $1.0\times 10^{-4}$ J.
At angular frequency $\omega = 1000$ rad/s, what is the impedance of a $2~\mu\text{F}$ capacitor?
Solution
For a capacitor,
Substitute the values:
So the impedance is $-j500~\Omega$.
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:
So
The other branch must carry $4$ A leaving the node.
Difficulty
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:
So
The current from $a$ to $b$ is $1$ A.
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:
The voltage source gives the constraint:
Substitute $V_a = V_b + 9$ into KCL:
Multiply by $6$:
Then
So $V_a = 12$ V.
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:
which simplifies to
Right mesh:
which simplifies to
From the second equation,
Substitute into the first:
Then
So $i_1 = 8$ A and $i_2 = 4$ A.
Two clockwise mesh currents $i_1$ and $i_2$ share a branch with a $1$ A current source.
The current source forces
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:
The current-source constraint is
From the constraint,
Substitute into the KVL equation:
Then
So $i_1 = 2$ A and $i_2 = 3$ A.
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:
Next, keep the $3$ A source and short the $12$ V source. Then the node sees two $6~\Omega$ resistors in parallel:
So the voltage contribution is
Add the contributions:
So the node voltage is $15$ V.
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:
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:
So the Thevenin equivalent is $V_{th} = 8$ V in series with $R_{th} = \frac{4}{3}~\Omega$.
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:
For a charging capacitor,
Here,
So
At $t = \tau$,
which is about $7.6$ V.
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:
So the total impedance is
Apply phasor Ohm's law:
Multiply by the complex conjugate:
In polar form, this is
Difficulty
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:
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:
So the load current is $1$ A.
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
So the load current is
The load power is
So the load absorbs $24$ W.
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:
The final current is
So the current is
At $t = 0.2$ s,
which is about $1.73$ A.
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
Real power is
So the load has real power about $416$ W and power factor $0.866$ lagging.
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,
Substitute the values:
So the output is $-2$ V.
Difficulty
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
So the best load is
The maximum power is
So $R_L = 5~\Omega$ and the maximum load power is $20$ W.
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
The final value is
The time constant is
So
or
To find when $v_C(t) = 12$ V:
So the capacitor reaches $12$ V after $0.2\ln 2 \approx 0.139$ s.
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,
Substitute the values:
Convert to hertz:
So the resonant frequency is about $50.3$ Hz.
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,
Substitute the values:
So the output voltage is $-8$ V.
