Set 1 — Fundamentals of Electric Circuits
Twenty-nine fully worked problems on the quantities every later chapter takes for granted: charge and current, voltage, power and energy, resistance and resistivity, and the average and RMS values of waveforms that are not constant. Nothing here needs a theorem — only the definitions, applied carefully and with the units watched.
Charge and current are a calculus pair. \(i = dq/dt\) going one way, \(q = \int i\,dt\) going the other. On a graph this means current is the slope of a \(q\!-\!t\) curve and charge is the area under an \(i\!-\!t\) curve.
The passive sign convention decides every sign. If current enters the terminal marked \(+\), then \(p = vi\) and a positive answer means the element absorbs. If current enters the \(-\) terminal, \(p = -vi\). A negative power means the element is delivering.
Power sums to zero. \(\sum p = 0\) over every element at every instant. Use it as a check on every solved circuit — it costs one line and catches most arithmetic slips.
Resistance is geometry times material. \(R = \rho L / A\), and \(R_T = R_0\left[1 + \alpha_0 (T - T_0)\right]\) for the temperature dependence.
Heating depends on the RMS value, never the average. \(P = I_{\mathrm{rms}}^{2}R\). Compute \(I_{\mathrm{rms}} = \sqrt{\tfrac{1}{T}\int_0^T i^2\,dt}\); the average value tells you about charge transfer, not heat.
The figure shows the current flowing into a capacitor. Determine the charge acquired by the capacitor during the first \(5\,\mu\text{s}\).
Since \(i = dq/dt\), the charge delivered is the area under the \(i\!-\!t\) curve. Split the interval at every break point of the waveform — here at \(t = 1,\,3\) and \(4\ \mu\text{s}\) — so that each piece is a triangle, a rectangle or a trapezium.
Evaluating the four areas, with time in \(\mu\text{s}\) and current in A so that the product is in \(\mu\text{C}\):
Adding them:
Two coils connected in parallel across a 100 V DC supply draw 10 A from the supply. The power dissipated in one coil is 600 W. Find the resistance of each coil.
The effective (parallel) resistance seen by the supply:
Because the coils are in parallel, the full supply voltage appears across each. Using the power in the first coil:
The total power drawn is \(P = VI = 100 \times 10 = 1000\ \text{W}\), so the second coil dissipates the remainder:
Confirming with the parallel formula:
How many 200 W / 220 V incandescent lamps connected in series would consume the same total power as a single 100 W / 220 V incandescent lamp? Assume the filament resistance stays constant.
A lamp's rating fixes its resistance, not its power: the marked power is what it draws at its marked voltage. Resistance of one 200 W lamp:
Resistance required to dissipate 100 W at 220 V:
Number of 242 Ω lamps in series needed to reach 484 Ω:
When a resistor \(R\) is connected to a current source it dissipates 18 W. When the same \(R\) is connected to a voltage source whose numerical magnitude equals that of the current source, it absorbs 4.5 W. Find the magnitude of the source and the value of \(R\).
With a current source of value \(I\) driving \(R\), all of the source current passes through the resistor:
The voltage source has the same numerical magnitude, so \(V = I\) as a number and
Multiplying the two equations eliminates \(R\) entirely:
Verification with the voltage source: \(P_2 = 3^2/2 = 4.5\ \text{W}\;\checkmark\)
The charge \(q(t)\) delivered by a constant-voltage source is shown. Determine the current supplied by the source at (a) \(t = 1\,\text{s}\) and (b) \(t = 3\,\text{s}\).
The current is the slope of the charge waveform, \(i = dq/dt\). Both requested instants fall in the interior of a straight segment, so the slope is simply \(\Delta q/\Delta t\) for that segment.
aSegment A→B, valid for \(0 \le t \le 2\,\text{s}\):
bSegment B→C, valid for \(2 \le t \le 5\,\text{s}\):
A toaster rated 1000 W, 240 V is connected to a 220 V supply. Will the toaster be damaged? Will its power rating be affected?
The rating fixes the element resistance and the maximum safe current:
Current actually drawn at 220 V:
Since \(3.82\,\text{A} < 4.167\,\text{A}\), the element is not damaged. The power actually consumed is
Equivalently, since \(P \propto V^2\) at fixed resistance:
What is the maximum voltage that can be applied across the series combination of a 150 Ω, 2 W resistor and a 100 Ω, 1 W resistor without exceeding the power rating of either resistor?
In series both resistors carry the same current, so work in terms of current. From \(P = I^2 R\) the maximum safe current of each is \(I_{\max} = \sqrt{P/R}\):
The combination is limited by whichever resistor saturates first — the smaller of the two currents, \(I = 0.100\,\text{A}\), set by the 100 Ω resistor:
At that current the 150 Ω resistor dissipates only
so it is running comfortably below its rating while the 100 Ω resistor is exactly at its limit.
A wire 50 m long and 2 mm² in cross-section has a resistance of 0.56 Ω. A 100 m length of the same material has a resistance of 2 Ω at the same temperature. Find the diameter of this second wire.
Both wires share the same material, so their resistivities are equal. Writing \(\rho = RA/L\) for each and equating:
Solving for the unknown area:
Note that \(A_1\) was left in mm², so \(A_2\) comes out in mm² too — the units cancel provided you are consistent, and there is no need to convert to m².
From \(A = \pi d^2/4\):
The voltage across a 10 Ω resistor is the square wave shown, of amplitude \(\pm 10\,\text{V}\) and period \(\pi\). Find the average power dissipated by the resistor.
Average power is \(P_{\text{avg}} = V_{\text{rms}}^2/R\), with \(V_{\text{rms}}^2 = \tfrac{1}{T}\int_0^T v^2\,dt\). Because power depends on \(v^2\), the sign of the voltage is irrelevant — both parts of the cycle contribute positively:
Hence \(V_{\text{rms}} = 10\ \text{V}\) and
A wire-wound resistor is to be made from 0.2 mm diameter constantan wire wound around a cylinder of 1 cm diameter. How many turns are needed for a resistance of 50 Ω? Take the resistivity of constantan as \(49 \times 10^{-8}\ \Omega\text{m}\).
Cross-sectional area of the wire:
Required wire length from \(R = \rho L / A\):
Each turn wraps once around the cylinder, so its length is the circumference \(2\pi r\) with \(r = 0.5\,\text{cm} = 0.005\,\text{m}\):
Find the average value of the periodic current waveform shown, of period \(\pi\) s, which rises linearly from 5 A to 10 A and falls linearly back to 5 A.
Describe the waveform piecewise over one period:
Check the end points: at \(t = \pi/2\) both expressions give 10 A, and at \(t = \pi\) the second gives 5 A — the waveform is continuous and periodic.
The average value over one period is \(I_{\text{avg}} = \tfrac{1}{T}\int_0^{T} i\,dt\):
The domestic power load in a house comprises eight 100 W lamps, three 80 W fans, one ½ hp refrigerator and one 1000 W heater.
- Calculate the total current taken from the 230 V supply if everything is switched on.
- Calculate the energy consumed in a day if, on average, only a quarter of the load is on at any time.
Tabulating the connected load, with 1 hp = 746 W:
| Item | Quantity | Load |
|---|---|---|
| Lamps | 8 × 100 W | \(800\ \text{W}\) |
| Fans | 3 × 80 W | \(240\ \text{W}\) |
| Refrigerator | 1 × ½ hp | \(0.5 \times 746 = 373\ \text{W}\) |
| Heater | 1 × 1000 W | \(1000\ \text{W}\) |
| Total connected load | \(2413\ \text{W}\) |
aTotal current from the supply:
bEnergy per day with a quarter of the load on continuously over 24 h:
The figure shows the current through a practical inductor of resistance 1 Ω and inductance 2 H. The current ramps linearly from 0 to 6 A during the first 2 s and then stays at 6 A. Find the total energy absorbed by the inductor in the first four seconds.
A practical inductor is modelled as an ideal 2 H inductance in series with a 1 Ω resistance. Energy goes to two different places, and they must be computed separately.
Energy stored in the magnetic field, which depends only on the end points because \(w_L = \tfrac{1}{2}Li^2\) is a state function:
Energy dissipated in the 1 Ω resistance, which depends on the whole history. For \(0\le t\le 2\), \(i = 3t\); for \(2\le t\le 4\), \(i = 6\):
Total energy absorbed by the physical component:
The current entering the positive terminal of a device is \(i(t) = 3e^{-2t}\ \text{A}\) and the voltage across it is \(v(t) = 5\,\dfrac{di}{dt}\ \text{V}\). Find (a) the charge delivered between \(t = 0\) and \(t = 2\,\text{s}\), (b) the power absorbed, and (c) the energy absorbed in 3 s.
aCharge is the integral of current:
bDifferentiate to obtain the voltage, then form the product:
The current enters the positive terminal, so the passive sign convention holds and \(p = vi\) is used directly with no extra minus sign.
cEnergy absorbed over the first 3 s:
The charge entering the positive terminal of an element is \(q = 10\sin 4\pi t\ \text{mC}\), while the voltage across it (plus to minus) is \(v = 2\cos 4\pi t\ \text{V}\). Find (a) the power delivered to the element at \(t = 0.3\,\text{s}\), and (b) the energy delivered between 0 and 0.6 s.
First recover the current by differentiating the charge:
aPower:
At \(t = 0.3\,\text{s}\) the argument is \(4\pi(0.3) = 1.2\pi\) rad, and \(\cos 1.2\pi = -0.809\), so \(\cos^2 = 0.6545\):
bEnergy, using \(\cos^2\theta = \tfrac{1}{2}(1+\cos 2\theta)\):
A copper field coil has a resistance of 50 Ω at 20 °C. The temperature coefficient of resistance of copper at 20 °C is \(\alpha_{20} = 0.00393\ /^{\circ}\text{C}\). Find its resistance when the coil heats up to 70 °C in service.
Resistance varies linearly with temperature. Because \(\alpha\) is quoted at 20 °C, 20 °C must be used as the reference:
A current rises linearly from 0 to 10 A over 2 s, drops instantly to 0, and repeats — a sawtooth of period 2 s. Find (a) its average value, (b) its RMS value, and (c) the power it delivers to a 4 Ω resistor.
Over one period the current is \(i(t) = 5t\) for \(0 \le t \le 2\,\text{s}\).
aAverage value:
bRMS value:
cHeating is governed by the RMS value:
A 12 V battery is connected to a 6 Ω resistor in series with the parallel combination of a 4 Ω and a 12 Ω resistor. Find (a) the total current drawn from the battery and (b) the power dissipated in the 4 Ω resistor.
aReduce the parallel pair first, working from the far end of the network back towards the source:
Total resistance and battery current:
bVoltage across the parallel section, then the 4 Ω branch power:
Power check across the whole circuit:
A household runs a 2 kW air-conditioner for 8 hours a day and ten 15 W LED bulbs for 6 hours a day. If electricity costs ₹7.50 per kWh, estimate the monthly (30-day) energy bill.
Daily energy of each load, taking care to express every power in kW before multiplying by hours:
Monthly energy and cost:
A 100 μF capacitor is charged to 200 V. (a) Find the charge and energy stored. (b) This capacitor is then connected across an identical, initially uncharged 100 μF capacitor. Find the final common voltage and the energy lost in the process.
aInitial charge and stored energy:
bCharge is conserved — it cannot leave the isolated pair — and is now shared over a total capacitance \(C_{\text{tot}} = 200\,\mu\text{F}\):
Final stored energy and the energy lost:
Further Problems
Nine additional problems covering ground the set above leaves out — conduction at the electron level, differentiation of a product, power balance when a dependent source is present, composite conductors, extraction of the temperature coefficient from measurements, and the RMS values of offset and rectified waveforms. The last two are genuine challenge problems that reward a general argument over arithmetic.
A copper conductor of cross-sectional area 2 mm² carries a steady current of 10 A. Copper contains \(n = 8.5 \times 10^{28}\) free electrons per cubic metre. Find (a) the current density, (b) the drift velocity of the electrons, and (c) the number of electrons crossing any section per second.
aCurrent density is current per unit area:
bIn time \(\Delta t\) the electrons within a length \(v_d\Delta t\) all cross the section, so \(I = n\,e\,A\,v_d\) and
That is 0.367 mm/s — an electron would take about 45 minutes to travel one metre.
cElectrons per second:
The total charge entering a terminal is \(q = 5t\sin 4\pi t\ \text{mC}\). Calculate the current at \(t = 0.5\,\text{s}\).
The charge is a product of two functions of time, so the product rule is needed:
Substituting \(t = 0.5\,\text{s}\), where \(4\pi t = 2\pi\) so \(\sin 2\pi = 0\) and \(\cos 2\pi = 1\):
In the single-loop circuit shown, the dependent source is a current-controlled voltage source of value \(2I\) volts. Find the loop current \(I\) and the power associated with each of the three elements, and verify conservation of power.
Apply KVL clockwise around the loop, taking a rise across the 24 V source and drops across the resistor and the dependent source:
Note that a dependent source is handled in KVL exactly like an independent one — the only difference is that its value must later be expressed in terms of the circuit unknowns, which here it already is.
Now evaluate each element under the passive sign convention. For the 24 V source, current leaves its positive terminal, so it supplies:
The resistor always absorbs:
The dependent source has terminal voltage \(v = 2I = 8\ \text{V}\), and the current enters its positive terminal, so it absorbs:
Conservation of power:
An aluminium wire 100 m long and 2 mm in diameter is joined end to end with a copper wire 50 m long and 1.5 mm in diameter. Take \(\rho_{\text{Al}} = 2.83\times10^{-8}\ \Omega\text{m}\) and \(\rho_{\text{Cu}} = 1.72\times10^{-8}\ \Omega\text{m}\). Find the resistance of the composite conductor, and the fraction of the total power lost in each section.
Cross-sectional areas:
Resistance of each section from \(R = \rho L/A\):
The two sections are in series, so the resistances add:
The same current flows through both, so power divides in the ratio of the resistances:
A coil measures 40 Ω at 20 °C and 48 Ω at 70 °C. Find (a) the temperature coefficient of resistance referred to 0 °C, (b) the resistance at 0 °C, (c) the resistance at 100 °C, and (d) the inferred temperature of zero resistance.
Referred to 0 °C the resistance law is \(R_T = R_0(1 + \alpha_0 T)\). Writing it at both measured temperatures:
aDividing the second by the first eliminates \(R_0\):
bBack-substituting:
cResistance at 100 °C:
dThe straight line \(R_T = R_0(1+\alpha_0 T)\) reaches zero when \(1 + \alpha_0 T = 0\):
A current \(i(t) = 4 + 6\sin\omega t\) A flows through a 5 Ω resistor. Find (a) the average value, (b) the RMS value, and (c) the average power dissipated.
aThe sine averages to zero over a whole number of cycles, so only the DC term survives:
bFor the RMS value, square first:
Averaging term by term: the constant gives 16, the cross term averages to zero over a whole cycle, and the mean square of a sine is \(\tfrac{1}{2}\), so the last term contributes \(36 \times \tfrac{1}{2} = 18\).
cAverage power:
A half-wave rectified sinusoidal current has a peak value of 10 A: it follows \(10\sin\omega t\) for the first half-cycle and is zero for the second. Find its average value, RMS value, form factor and peak factor.
Work in terms of \(\theta = \omega t\) with a period of \(2\pi\). The waveform is \(I_m\sin\theta\) on \((0,\pi)\) and zero on \((\pi,2\pi)\).
Average value — note that the denominator is the full period, even though the current is non-zero for only half of it:
RMS value, using \(\sin^2\theta = \tfrac{1}{2}(1-\cos2\theta)\):
Form factor and peak factor:
Four identical resistors, each 100 Ω and rated 1 W, are available. Find the maximum voltage that may be applied, and the maximum total power the combination can safely dissipate, when they are connected (a) all four in series, (b) all four in parallel, and (c) as two parallel branches of two series resistors. Comment on the pattern.
Each resistor is safe up to \(I_{\max} = \sqrt{P/R} = \sqrt{1/100} = 0.1\ \text{A}\), equivalently \(V_{\max} = \sqrt{PR} = 10\ \text{V}\) across it.
aAll in series. Equivalent resistance \(400\ \Omega\), and all four carry the same current, which is limited to 0.1 A:
bAll in parallel. Equivalent resistance \(25\ \Omega\), and all four see the same voltage, limited to 10 V:
cTwo-by-two. Each branch is \(200\ \Omega\); the two in parallel give \(R_{\text{eq}} = 100\ \Omega\). Each branch current is limited to 0.1 A, so the branch voltage — which is the applied voltage — is
The maximum applied voltage differs by a factor of four across the three arrangements, yet the safe dissipation is 4 W in every case.
An uncharged capacitor \(C\) is charged from a DC source of voltage \(V\) through a series resistance \(R\), until it is fully charged. Find the energy drawn from the source, the energy finally stored, and the energy dissipated in \(R\). Show that the efficiency of the process is 50 % regardless of the value of \(R\). Evaluate for \(C = 1000\ \mu\text{F}\) and \(V = 100\ \text{V}\).
When charging is complete the capacitor holds \(Q = CV\). Every coulomb of that charge was pushed through the source, and the source terminal voltage was \(V\) throughout, so
Notice that this argument never mentions \(R\) — it does not even require the current waveform to be known.
The energy finally stored in the capacitor is the standard result:
Energy is conserved, so whatever the source delivered and the capacitor did not keep must have gone into the resistor:
Hence the efficiency:
Direct confirmation by integration, using \(i = (V/R)e^{-t/RC}\):
The \(R\) in the numerator cancels the \(R\) in the time constant exactly — halving the resistance doubles the peak current but halves the duration, leaving the integral unchanged.
Numerically, with \(C = 1000\ \mu\text{F}\) and \(V = 100\ \text{V}\):
Practice Problems
Work each of these on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not, because producing it is the exercise.
P1. The charge entering a terminal is \(q = (5t^2 + 3t)\ \text{mC}\). Find the current at \(t = 2\,\text{s}\), and the charge delivered between \(t = 1\) and \(t = 3\,\text{s}\).
Show answer
\(i = (10t+3)\) mA, so \(i(2) = 23\) mA. \(\Delta q = q(3)-q(1) = 54 - 8 = 46\) mC.P2. A 60 W, 120 V lamp and a 100 W, 120 V lamp are connected in series across a 120 V supply. Which lamp glows brighter, and what power does each dissipate?
Show answer
\(R_{60} = 240\,\Omega\), \(R_{100} = 144\,\Omega\), total 384 Ω, \(I = 0.3125\) A. Then \(P_{60} = 23.4\) W and \(P_{100} = 14.1\) W — the 60 W lamp is brighter, because in series the larger resistance takes the larger share.P3. The current through a 5 Ω resistor is \(i = 4e^{-3t}\) A. Find the total energy dissipated as \(t \to \infty\).
Show answer
\(W = \int_0^{\infty}80e^{-6t}dt = 80/6 = 13.33\) J.P4. A copper busbar of rectangular section 40 mm × 6 mm and length 3 m carries 800 A. Take \(\rho = 1.72\times10^{-8}\ \Omega\text{m}\). Find its resistance, the voltage drop and the power lost.
Show answer
\(A = 240\ \text{mm}^2\), \(R = 215\ \mu\Omega\), \(V = 0.172\) V, \(P = 137.6\) W.P5. An element has \(v = 8\) V across it with the \(+\) mark on the left, and a current of 3 A flows into the right-hand terminal. Is it absorbing or supplying, and how much?
Show answer
Current enters the \(-\) terminal, so the passive sign convention is violated and \(p = -vi = -24\) W. The element supplies 24 W.P6. A triangular current waveform rises linearly from 0 to \(I_m\) in \(T/2\) and falls linearly back to 0 in the remaining \(T/2\). Derive its average and RMS values, and its form factor.
Show answer
\(I_{\text{avg}} = I_m/2\), \(I_{\text{rms}} = I_m/\sqrt3\), form factor \(2/\sqrt3 = 1.155\) — identical to the sawtooth of Problem 17, since squaring destroys the distinction between rising and falling.P7. A battery of emf 12 V and internal resistance 0.5 Ω supplies a 5.5 Ω load. Find the terminal voltage, the load power, and the efficiency of energy transfer.
Show answer
\(I = 2\) A, \(V_T = 11\) V, \(P_L = 22\) W, \(\eta = 5.5/6 = 91.7\,\%\).P8. A tungsten filament has a resistance of 20 Ω at 20 °C and 240 Ω at its operating temperature. Take \(\alpha_{20} = 0.0045\,/^{\circ}\text{C}\). Estimate the operating temperature.
Show answer
\(240 = 20[1+0.0045(T-20)]\) gives \(T - 20 = 2444\), so \(T \approx 2464\,^{\circ}\text{C}\) — close to the real figure for an incandescent lamp, which is why cold inrush current is roughly twelve times the running current.P9. A 250 V DC motor draws 15 A and delivers 3.2 kW of mechanical power. Find its efficiency and the energy wasted in one 8-hour shift.
Show answer
Input \(= 3750\) W, \(\eta = 3200/3750 = 85.3\,\%\). Loss \(= 550\) W, so 4.4 kWh per shift.P10. Three resistors of 6 Ω, 12 Ω and 4 Ω are connected in parallel across 24 V. Find the total current and the power in each.
Show answer
\(R_{\text{eq}} = 2\ \Omega\), \(I = 12\) A. Powers: 96 W, 48 W, 144 W; total 288 W \(= 24 \times 12\;\checkmark\)P11. A current \(i = 10\sin 100\pi t\) A flows into a terminal. How much charge passes in the first half-cycle, and what is the net charge over a full cycle?
Show answer
Half-cycle: \(q = \int_0^{0.01}10\sin100\pi t\,dt = 20/(100\pi) = 63.7\) mC. Over a full cycle the net charge is zero — which is precisely what "alternating" means.P12. Two wires of the same material and the same mass have lengths in the ratio 1 : 2. Find the ratio of their resistances.
Show answer
Equal mass and density means equal volume, so \(A \propto 1/L\) and therefore \(R = \rho L/A \propto L^2\). The ratio is 1 : 4.
Challenge Problems
Each of these needs an idea rather than a formula. Do not open the answer until you have at least decided what the governing principle is.
C1. An infinite ladder is built from identical resistors: a resistor \(R\) in series, then \(R\) in shunt, then \(R\) in series, and so on for ever. Find the input resistance \(R_{\text{in}}\) seen at the near end.
Show answer
Because the ladder is infinite, removing the first two resistors leaves an identical ladder. Hence \(R_{\text{in}} = R + (R \parallel R_{\text{in}})\), giving \(R_{\text{in}}^2 - RR_{\text{in}} - R^2 = 0\) andthe golden ratio times \(R\). The negative root is discarded as unphysical.\[ R_{\text{in}} = \frac{R\left(1+\sqrt5\right)}{2} = 1.618R \]C2. A resistor of resistance \(R\) is to be connected across a source of emf \(E\) with internal resistance \(r\). Two students disagree: one says \(R = r\) is best because it maximises the power in the load; the other says \(R \gg r\) is best because it maximises efficiency. Reconcile them, and state which criterion applies to (i) a power transmission line and (ii) a radio receiving antenna.
Show answer
Both are right about different objectives. Load power \(P_L = E^2R/(R+r)^2\) peaks at \(R=r\), where \(P_L = E^2/4r\) and efficiency is only 50 %. Efficiency \(\eta = R/(R+r)\) rises monotonically towards 100 % as \(R\to\infty\), but the delivered power then falls to zero.
(i) A transmission line carries abundant power and pays for every wasted joule, so efficiency governs: \(R \gg r\), which is why generators have low internal impedance and lines are run at high voltage.
(ii) An antenna receives a signal of fixed, tiny power that cannot be increased, so extracting the maximum of it matters and the 50 % loss is irrelevant: match, \(R = r\).C3. A resistive heater is fed from a 230 V, 50 Hz supply through a phase-controlled switch, which blocks the first \(\alpha\) radians of every half-cycle and conducts for the rest. Derive an expression for the RMS voltage across the heater as a function of \(\alpha\), and hence find the firing angle that gives exactly half the rated heat output.
Show answer
With \(v = V_m\sin\theta\) for \(\alpha \le \theta \le \pi\) and zero before \(\alpha\), and averaging over the half-period \(\pi\):Setting \(\alpha = 0\) recovers \(V_m/\sqrt2\), as it must. Half the heat means half the value of \(V_{\text{rms}}^2\), so\[ V_{\text{rms}}^{2} = \frac{1}{\pi}\int_{\alpha}^{\pi} V_m^2\sin^2\theta\,d\theta = \frac{V_m^{2}}{2\pi}\left[(\pi - \alpha) + \frac{\sin 2\alpha}{2}\right] \]This transcendental equation normally needs iteration, but here \(\alpha = \pi/2\) satisfies it exactly, since \(\sin\pi = 0\). So firing at 90° halves the power, and the heater sees \(V_{\text{rms}} = V_m/2 = 162.6\ \text{V}\).\[ \frac{(\pi-\alpha) + \tfrac{1}{2}\sin 2\alpha}{2\pi} = \frac{1}{4} \;\Longrightarrow\; \alpha - \tfrac{1}{2}\sin 2\alpha = \frac{\pi}{2} \]
The trap: 90° is halfway through the half-cycle, so it is tempting to think any symmetric argument gives half power — but that is a coincidence of the sine, not a general rule. Halving the conduction angle to 90° does not generally halve the power, and the RMS voltage falls only to 71 % of full, not 50 %, because power goes as the square.
Multiple-Choice Questions
MCQ 1. The area under an \(i\!-\!t\) curve represents:
(a) power (b) energy (c) charge (d) voltageShow answer
(c) charge. Since \(i = dq/dt\), integrating current gives charge. The area under a \(p\!-\!t\) curve would give energy.MCQ 2. Two resistors of 4 Ω and 6 Ω are connected in parallel across a source. The ratio of the power dissipated in the 4 Ω to that in the 6 Ω is:
(a) 2 : 3 (b) 3 : 2 (c) 4 : 9 (d) 9 : 4Show answer
(b) 3 : 2. In parallel the voltage is common, so \(P = V^2/R \propto 1/R\), giving \(1/4 : 1/6 = 3:2\). Option (a) is the answer for a series connection.MCQ 3. The RMS value of the current \(i = 3 + 4\sin\omega t\) A is:
(a) 5 A (b) 5.83 A (c) 4.12 A (d) 7 AShow answer
(c) 4.12 A. \(I_{\text{rms}} = \sqrt{I_{\text{dc}}^2 + I_m^2/2} = \sqrt{9 + 8} = \sqrt{17} = 4.123\) A. Distractor (a) treats both 3 and 4 as peaks and adds them in quadrature; (d) adds them arithmetically; (b) is the answer to Problem 26, a different waveform.MCQ 4. An element carries 2 A into its negative terminal with 6 V across it. The element:
(a) absorbs 12 W (b) supplies 12 W (c) absorbs 3 W (d) supplies 3 WShow answer
(b) supplies 12 W. The passive sign convention is violated, so \(p = -vi = -12\) W, meaning 12 W is delivered.MCQ 5. Two wires of the same material have the same length, but one has twice the diameter of the other. The ratio of their resistances (thin : thick) is:
(a) 2 : 1 (b) 4 : 1 (c) 1 : 2 (d) 1 : 4Show answer
(b) 4 : 1. \(R \propto 1/A \propto 1/d^2\), so doubling the diameter quarters the resistance.MCQ 6. The form factor of a half-wave rectified sine wave is:
(a) 1.11 (b) 1.155 (c) 1.571 (d) 2.0Show answer
(c) 1.571 \(= \pi/2\). Option (a) is the full sine wave, (b) the triangular or sawtooth wave, and (d) is the peak factor of the half-wave rectified sine.MCQ 7. A 100 W, 250 V lamp is operated at 200 V. Assuming constant resistance, the power consumed is:
(a) 80 W (b) 64 W (c) 100 W (d) 125 WShow answer
(b) 64 W. At fixed \(R\), \(P \propto V^2\), so \(100 \times (200/250)^2 = 100 \times 0.64\). Option (a) is the trap of scaling linearly.MCQ 8. The temperature coefficient of resistance of a conductor:
(a) is independent of the reference temperature (b) decreases as the reference temperature rises (c) increases as the reference temperature rises (d) is always negative for metalsShow answer
(b) decreases as the reference temperature rises. Since \(\alpha_T = \alpha_0/(1+\alpha_0 T)\), quoting \(\alpha\) without its reference temperature is meaningless. Metals have positive \(\alpha\); semiconductors and carbon are negative.MCQ 9. Charging a capacitor from a DC voltage source through a resistor has a maximum theoretical efficiency of:
(a) 100 % (b) 75 % (c) 50 % (d) it depends on \(R\)Show answer
(c) 50 %, independent of \(R\) — see Problem 29. Reducing \(R\) shortens the transient but raises the current in exactly compensating proportion.MCQ 10. One kilowatt-hour is equal to:
(a) 1000 J (b) 3600 J (c) \(3.6\times10^{6}\) J (d) \(10^{6}\) JShow answer
(c) \(3.6\times10^{6}\) J. Option (b) is one watt-hour.MCQ 11. The drift velocity of electrons in a copper conductor carrying a normal current is of the order of:
(a) \(10^{8}\) m/s (b) \(10^{3}\) m/s (c) \(10^{-4}\) m/s (d) \(10^{-12}\) m/sShow answer
(c) about \(10^{-4}\) m/s — a fraction of a millimetre per second (Problem 21). Option (a) is the speed of the electric field's propagation, which is what makes the lamp light instantly.MCQ 12. Four 100 Ω, 1 W resistors are connected in series. The maximum power the combination can safely dissipate is:
(a) 1 W (b) 2 W (c) 4 W (d) 16 WShow answer
(c) 4 W. Each resistor may dissipate its rated 1 W, and they all reach that limit together because the arrangement is symmetric — see Problem 28. The voltage limit changes with topology; the power limit does not.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Current | \(i = \dfrac{dq}{dt}\) | Slope of the \(q\!-\!t\) curve |
| Charge | \(q = \displaystyle\int_{t_0}^{t} i\,dt\) | Area under the \(i\!-\!t\) curve |
| Voltage | \(v = \dfrac{dw}{dq}\) | Energy per unit charge; 1 V = 1 J/C |
| Power | \(p = vi = i^2R = \dfrac{v^2}{R}\) | \(+\) absorbed under the passive sign convention |
| Energy | \(w = \displaystyle\int p\,dt\) | 1 kWh \(= 3.6\times10^6\) J |
| Conservation of power | \(\sum p = 0\) | Holds at every instant, not just on average |
| Ohm's law | \(v = iR\) | Conductance \(G = 1/R\) in siemens |
| Resistance | \(R = \dfrac{\rho L}{A}\) | \(\rho_{\text{Cu}} = 1.72\times10^{-8}\ \Omega\text{m}\) |
| Temperature effect | \(R_T = R_0\!\left[1 + \alpha_0(T-T_0)\right]\) | \(\alpha_T = \alpha_0/(1+\alpha_0T)\) |
| Series / parallel | \(R_s = \sum R_k\), \(\dfrac{1}{R_p} = \sum\dfrac{1}{R_k}\) | Two in parallel: \(R_1R_2/(R_1+R_2)\) |
| Average value | \(X_{\text{avg}} = \dfrac{1}{T}\displaystyle\int_0^T x\,dt\) | Governs charge transfer, electroplating |
| RMS value | \(X_{\text{rms}} = \sqrt{\dfrac{1}{T}\displaystyle\int_0^T x^2\,dt}\) | Governs heating; \(P = I_{\text{rms}}^2R\) |
| DC + AC combination | \(I_{\text{rms}} = \sqrt{I_{\text{dc}}^2 + I_{\text{ac,rms}}^2}\) | Quadrature, never arithmetic |
| Form / peak factor | \(k_f = \dfrac{X_{\text{rms}}}{X_{\text{avg}}}\), \(k_p = \dfrac{X_m}{X_{\text{rms}}}\) | Sine: 1.11, 1.414 · Sawtooth: 1.155, 1.732 · Half-wave: 1.571, 2 |
| Current density | \(J = \dfrac{I}{A} = n e v_d\) | \(n_{\text{Cu}} = 8.5\times10^{28}\ \text{m}^{-3}\) |
| Stored energy | \(w_C = \tfrac{1}{2}Cv^2\), \(w_L = \tfrac{1}{2}Li^2\) | Recoverable; resistive loss is not |
Common Mistakes
Decomposing a graph into pieces whose widths do not add up. When integrating an \(i\!-\!t\) curve graphically, always check that the sub-interval widths sum to the total interval. This is the error that corrupted the original version of Problem 1.
Using the average value to compute power. Heating depends on \(I_{\text{rms}}^2R\). For the sawtooth of Problem 17 the average gives 100 W against a true 133.3 W — a 25 % error.
Adding DC and AC RMS values arithmetically. They combine in quadrature. For \(4 + 6\sin\omega t\) the answer is 5.83 A, not 8.24 A.
Adding power ratings. A 2 W and a 1 W resistor in series are not a 3 W combination — the weaker component sets the limit. This differs from the symmetric case of Problem 28, where identical resistors do reach their limits together.
Scaling power linearly with voltage. At fixed resistance \(P \propto V^2\). A 20 % voltage drop gives a 36 % power drop, not 20 %.
Quoting \(\alpha\) without its reference temperature. \(\alpha_{20} = 0.00393\) and \(\alpha_0 = 0.00427\) describe the same copper. Substituting one into a formula written for the other is a routine source of a few per cent of error.
Forgetting the product rule. Differentiating \(q = 5t\sin4\pi t\) as though the \(5t\) were constant loses a whole term — see Problem 22.
Assuming an active element always supplies. Dependent sources routinely absorb power, as in Problem 23, and a battery being charged absorbs. "Active" means capable of supplying, not guaranteed to.
Dividing by the wrong period for a rectified waveform. A half-wave rectified sine is non-zero for only half the cycle, but the average is still taken over the full period \(2\pi\). Dividing by \(\pi\) doubles the answer.
Losing a factor of a thousand. Mixing mA with A, or mm² with m², is by a wide margin the most frequent error in this entire set. Carry the units through every line rather than reinstating them at the end.
Everything here has come from definitions applied to one element at a time, or to combinations simple enough to reduce by inspection. The next set pushes series–parallel reduction as far as it will go, and then meets the networks where it fails altogether — the bridge, where no two resistors are in series or in parallel — and introduces the delta–wye transformation that rescues them.
Before moving on, make sure the passive sign convention is automatic and that you never reach for an average value when a question asks about heating. Those two habits alone will carry you through most of Part 1.