Free FE Electrical Practice Problems with Step-by-Step Solutions

Answer: D) 4 A

Both resistors are connected between the same two nodes, so both see the whole 24 V and the source current is simply the sum of the two branch currents:

$$I_1 = \dfrac{24}{8} = 3\text{ A}, \qquad I_2 = \dfrac{24}{24} = 1\text{ A}, \qquad I_T = 4\text{ A}$$

The equivalent resistance gives the same thing: \(R_P = (8)(24)/(8+24) = 6\;\Omega\) and \(24/6 = 4\) A. Either route works, and it is worth being fluent in both, because branch currents are what you need for power in each resistor while the equivalent is what you need to see the circuit from the source.

3 A and 1 A are the two branch currents taken on their own. Whichever branch you stop at, the answer is short by the other one; the total current out of a source always exceeds any single branch it feeds.

0.75 A treats the two resistors as a series string, \(24/(8+24) = 24/32\). That would be right if the current had to pass through both, but the filled dots show the two branches meeting at both ends, so each carries its own current.

Answer: D) +2.31 × 10⁶ V/m

Three fields to add, and the only difficulty is the sign of each. The Handbook gives the magnitude of a point-charge field as

$$E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$$

and the direction follows one rule: the field points away from a positive charge and toward a negative one.

From q1, at a distance of 0.15 m and positive, so the field at P points away from it, in +x:

$$E_1 = \dfrac{2 \times 10^{-6}}{4\pi(8.85 \times 10^{-12})(0.15)^2} = 0.80 \times 10^{6}\ \text{V/m}$$

From q2, also 0.15 m away but negative, so the field at P points toward it, which from P is also +x. Twice the charge at the same distance gives twice the field:

$$E_2 = 1.60 \times 10^{6}\ \text{V/m}$$

From q3, 0.45 m away and positive, so the field points away from it, in −x. Three times the distance means a ninth of the field of an equal charge at 0.15 m:

$$E_3 = 0.089 \times 10^{6}\ \text{V/m}$$

$$E = 0.80 + 1.60 - 0.089 = +2.31 \times 10^{6}\ \text{V/m}$$

The two nearest charges push the same way, which is the point of the arrangement: q1 repels a test charge to the right and q2 attracts it to the right, so they add rather than cancel even though their signs differ. The far charge subtracts, but at three times the distance it removes only about four per cent.

−0.89 × 10⁶ V/m is the answer with the sign of q2's field reversed, that is, taking the field of a negative charge to point away from it. It is the single most common error here, and it flips the sum from +2.31 to −0.89. +1.60 × 10⁶ V/m is q2 alone, the largest of the three terms, and +0.80 × 10⁶ V/m is q1 alone; both are what happens when superposition is dropped and only one charge is counted.

Answer: B) \(6\,\text{V}\)

Handbook: FE Reference Handbook 10.6, Electrical and Computer Engineering → RC and RL Transients, p. 367 (PDF viewer: page 373), whose step response

$$v_C(t) = v_C(0)e^{-t/RC} + V\left(1 - e^{-t/RC}\right)$$

settles to \(V\), the driving voltage the capacitor sees, once the exponentials have died. The whole question is what voltage the capacitor sees.

Step 1 – A capacitor at DC steady state is an open circuit. Its current is \(C\,dv/dt\), and in steady state nothing changes, so \(dv/dt = 0\) and no current flows into it. Remove it from the diagram: what is left is \(R_1\) in series with \(R_2\) across the source.

Step 2 – The remaining circuit is a voltage divider. The same current flows through both resistors:

$$I = \frac{10}{R_1 + R_2} = \frac{10}{2{,}000 + 3{,}000} = 2\;\text{mA}$$

Step 3 – The capacitor is in parallel with \(R_2\), so it charges to the voltage across \(R_2\).

$$v_C(\infty) = V_s\left(\frac{R_2}{R_1 + R_2}\right) = 10\left(\frac{3{,}000}{5{,}000}\right) = 6\;\text{V}$$

or equivalently \(IR_2 = (2\;\text{mA})(3\;\text{k}\Omega) = 6\) V.

Answer: (B).

Check with KVL. The other 4 V must appear across \(R_1\): \(IR_1 = (2\;\text{mA})(2\;\text{k}\Omega) = 4\) V, and \(6 + 4 = 10\) V. Any answer that does not leave the remaining volts on the other resistor is wrong.

How long it takes to get there. Not asked, but worth seeing: the resistance the capacitor sees when the source is turned off is \(R_1 \parallel R_2 = 1.2\) kΩ, so \(\tau = (1{,}200)(0.5\,\mu\text{F}) = 0.6\) ms and the capacitor is within 1% of 6 V after about 3 ms. Note that the resistance setting the final value and the resistance setting the speed are different combinations of the same two resistors.

Step 4 – Eliminate the distractors:

(A) 10 V assumes the capacitor charges to the full source voltage. That happens only when nothing else drops voltage — with \(R_2\) drawing 2 mA through \(R_1\), four of the ten volts are lost before the capacitor is reached.

(C) 0 V confuses "blocks DC current" with "has no DC voltage". The capacitor does block the current, which is exactly why it ends up sitting at a steady voltage rather than at zero.

(D) 4 V is the divider taken the wrong way round, \(V_sR_1/(R_1+R_2)\). That is the voltage across \(R_1\), not across the parallel pair; it is the part of the source the capacitor does not get.

Answer: B) 2

Handbook: FE Reference Handbook 10.6, Mathematics → Discrete Math → Directed Graphs, or Digraphs, of Relation, p. 36 (PDF viewer: page 42). The definition is one sentence, and it is the whole question:

“A directed graph, or digraph, consists of a set \(V\) of vertices (or nodes) together with a set \(E\) of ordered pairs of elements of \(V\) called edges (or arcs). For edge \((a, b)\), the vertex \(a\) is called the initial vertex and vertex \(b\) is called the terminal vertex. An edge of form \((a, a)\) is called a loop.”

Step 1 – Write down the edge set from the drawing. Seven arrows:

$$E = \{(a,a),\; (a,b),\; (a,d),\; (b,c),\; (c,b),\; (c,a),\; (d,d)\}$$

Step 2 – Apply the definition, one edge at a time. A loop is a single edge whose initial and terminal vertex are the same, so the test is simply whether the pair repeats a letter:

$$(a,a)\ \checkmark \qquad (d,d)\ \checkmark \qquad \text{the other five have two different vertices}$$

Step 3 – Answer: 2 — (B).

A loop is not the same thing as a cycle. This graph also contains two closed routes: \(b \to c \to b\), using the pair of curved arrows, and \(a \to b \to c \to a\) around the outside. Both return to where they started, and both are easy to call “loops” in ordinary speech — but each of them is made of several edges. The Handbook's loop is one edge, drawn as an arrow that leaves a vertex and comes straight back to the same vertex. Only \(a\) and \(d\) carry one.

Step 4 – Eliminate the distractors:

(A) 1 counts only the loop at \(d\). The one at \(a\) is the easier of the two to miss, because \(a\) already has three other edges meeting it; count loops vertex by vertex rather than by scanning the picture.

(C) 3 is the two loops plus the \(b \to c \to b\) pair counted as a third. Two arrows between the same two vertices, one each way, are two ordinary edges — \((b,c)\) and \((c,b)\) — not a loop.

(D) 4 counts every closed route in the graph: the two loops, the \(b \to c \to b\) pair and the triangle \(a \to b \to c \to a\). That is the number of cycles, not the number of loops.

Answer: C) 10

Walk the twelve accesses, keeping the four lines in order of use with the least recently used one on the left.

The first five accesses are all new blocks, so 1, 4, 6 and 2 miss and fill the cache, with the repeat of 1 at step 3 hitting on the way. That is four misses and one hit, and the cache now holds 4, 6, 2 and 1, in that order of use.

From here every miss costs a block. Block 8 at step 6 misses and evicts 4, the least recently used. Block 1 at step 7 is still resident and hits, which also makes it the most recent. Block 3 at step 8 misses and evicts 6; block 6 at step 9 has just gone, so it misses too and evicts 2; block 9 at step 10 evicts 8; block 5 at step 11 evicts 1; and block 2 at step 12 evicts 3.

That is seven more misses on top of the first four, so ten of the twelve accesses miss and two hit.

8 is the trap, and it is the count you get by asking how many distinct blocks appear: 1, 2, 3, 4, 5, 6, 8 and 9, eight of them, each missing once the first time it is seen. That would be the whole story in a cache with eight lines, where nothing ever has to be evicted. This cache holds four, so six of the eight are thrown out and two of them, 6 and 2, are asked for again afterwards and miss a second time.

The other two numbers are what the same trace gives under two of the other replacement policies the Handbook lists. Replacing the most recently used block instead of the least recently used one leaves 9 misses: it keeps evicting the block just brought in, which happens to protect the old ones, so the repeat of 6 at step 9 survives. First-in, first-out gives 11, because it does not count a hit as a use: block 1 is refreshed by the hit at step 3 under least recently used but stays first in the queue under first-in, first-out, so it is evicted at step 6 and misses at step 7.

Answer: C) Periodic monitoring, medical surveillance and notification are triggered.

There are two numbers on the scale and the measurement lies between them. That position is the whole answer, and it is a position the regulation is built to recognise.

The permissible exposure limit is a ceiling: above it, the exposure has to be brought down. The action level, set here at 30, is a lower trip point that starts a programme rather than a control: once exposures reach it, the employer has to keep measuring, has to put the exposed workers under medical surveillance, and has to tell each worker what their own result was. At 35 the exposure is over the action level and under the limit, so the programme is triggered and no shutdown is.

The reason a lower trip point exists at all is worth understanding rather than memorising. Lead accumulates in the body; the harm is a function of dose summed over months, not of any single day. And the measurement is an eight-hour time-weighted average, so a compliant number can contain a much worse hour. A threshold that fires only at the limit would find the problem after it had happened.

The Handbook carries the same two-tier structure for a different hazard, on p. 35, and it is worth reading across: for occupational noise, a dose above 100% requires abatement, while a dose merely between 50% and 100% requires a hearing conservation programme. The upper number stops the exposure; the lower one starts the watching. Lead is laid out the same way.

Doing nothing reads the limit as the only number that matters, which is exactly the reading the action level exists to prevent. Stopping the soldering until the exposure is zero mistakes a limit for a prohibition; the standard sets a permissible exposure, not a forbidden substance, and no industrial process is run to zero. Respirators without surveillance inverts the order of the controls twice over: respiratory protection is the last resort, after ventilation and process change, and it is the surveillance that is mandatory here, not the respirator.