Electricity

This topic covers electric charge, current, potential difference, resistance, circuits, and Electromagnetic induction. Electricity is one of the core areas of the Leaving Certificate Physics Syllabus.

Electric Charge and Current

Explore the simulation above to develop intuition for this topic.

Electric Charge (OL/HL)

• Charge is measured in coulombs (C).
• The elementary charge e = 1.6 \times 10^{-19}\mathrm{ C.
• Like charges repel; unlike charges attract.

Electric Current (OL/HL)

$$I = \frac{Q}{t}$$

Where $I$ is current in amperes (A), $Q$ is charge in coulombs (C), and $t$ is time in seconds (s).

Current is the rate of flow of charge. Conventional current flows from positive to negative.

Example (OL): A charge of 60 C passes through a conductor in 2 minutes. Find the current.

I = \frac{60}{120} = 0.5\mathrm{ A

Current in Terms of Electron Flow (HL)

$$I = nAve$$

Where $n$ is the number density of free electrons, $A$ is the cross-sectional area, $v$ is the drift Velocity, and $e$ is the elementary charge.

Example (HL): A copper wire of cross-sectional area 2 \times 10^{-6}\mathrm{ m^2 carries a Current of 5 A. If n = 8.5 \times 10^{28}\mathrm{ m^{-3}Find the drift velocity.

V = \frac{I}{nAe} = \frac{5}{8.5 \times 10^{28} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}} \approx 1.84 \times 10^{-4}\mathrm{ m/s

Why Drift Velocity Is So Slow

Even a current of 1 A corresponds to about $6.25 \times 10^{18}$ electrons per second. The drift Velocity in a copper wire is less than 0.1 mm/s. The electric signal propagates at nearly The speed of light, but the individual electrons move extraordinarily slowly. This is analogous to People in a crowd: one person pushes, and the push propagates through the crowd quickly, but each Individual moves only a small step.

Potential Difference and EMF

Potential Difference (OL/HL)

$$V = \frac{W}{Q}$$

The potential difference (voltage) between two points is the energy transferred per unit charge. Unit: volt (V).

Electromotive Force (EMF) (OL/HL)

EMF is the energy supplied per unit charge by a source:

\mathrm{EMF = \frac{W}{Q}

Internal Resistance (HL)

A real cell has internal resistance $r$. The terminal voltage is:

V = \mathrm{EMF - Ir

Why Terminal PD Decreases with Current

The internal resistance dissipates energy as heat. As current increases, the internal voltage drop ($Ir$) increases, and the terminal PD decreases. If the battery is short-circuited ($R = 0$), all The EMF is dropped across the internal resistance, and the terminal PD is zero. The current is $I = \varepsilon / r$Which can be very large, potentially damaging the battery.

Resistance and Ohm’s Law

Ohm’s Law (OL/HL)

For a conductor at constant temperature:

$$V = IR$$

A conductor that obeys Ohm’s law has a constant resistance and is called an ohmic conductor.

Example (OL): A 12 V battery is connected to a resistor of 6 ohms. Find the current.

I = \frac{V}{R} = \frac{12}{6} = 2\mathrm{ A

Resistivity (OL/HL)

$$R = \frac{\rho L}{A}$$

Where $\rho$ is the resistivity (ohm-metre), $L$ is the length, and $A$ is the cross-sectional area.

Example (OL): A copper wire of length 100 m and cross-sectional area 1 \times 10^{-6}\mathrm{ m^2 has resistivity 1.7 \times 10^{-8}\mathrm{ ohm m. Find its Resistance.

R = \frac{1.7 \times 10^{-8} \times 100}{1 \times 10^{-6}} = 1.7\mathrm{ ohms

Why Resistivity Is Temperature-Dependent for Metals

In a metal, the ions vibrate more vigorously at higher temperatures, scattering conduction electrons More frequently. This increases resistivity. For semiconductors, resistivity decreases with Temperature as more charge carriers are liberated.

Series and Parallel Circuits (OL/HL)

Series Circuits

• Current is the same through all components: $I = I_1 = I_2 = \cdots$
• Total voltage equals the sum of individual voltages: $V = V_1 + V_2 + \cdots$
• Total resistance: $R = R_1 + R_2 + \cdots$

Parallel Circuits

• Voltage is the same across all branches: $V = V_1 = V_2 = \cdots$
• Total current equals the sum of branch currents: $I = I_1 + I_2 + \cdots$
• Total resistance: $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$

Example (OL): Find the total resistance of a 4 ohm and a 6 ohm resistor in parallel.

\frac{1}{R} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \implies R = 2.4\mathrm{ ohms

Why Adding Parallel Branches Reduces Total Resistance

Each additional branch provides an extra path for current. More paths means less total opposition. The total resistance is always less than the smallest individual resistance.

Kirchhoff’s Laws (HL)

Junction rule: The sum of currents entering a junction equals the sum leaving.

Loop rule: The sum of EMFs around any closed loop equals the sum of potential drops.

Applying Kirchhoff’s Laws Systematically

1. Label all currents and voltages.
2. Apply the junction rule at each junction.
3. Apply the loop rule to each independent loop.
4. Solve the resulting system of equations.

Example (HL): Find the current in each resistor for the following circuit: a 12 V battery in Series with a 2 ohm resistor, then a parallel combination of 6 ohm and 3 ohm resistors.

Parallel resistance: \frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2} \implies R_p = 2\mathrm{ ohms.

Total resistance: R = 2 + 2 = 4\mathrm{ ohms.

Total current: I = \frac{12}{4} = 3\mathrm{ A.

Voltage across parallel combination: V_p = 12 - 3 \times 2 = 6\mathrm{ V.

Current through 6 ohm: I_6 = \frac{6}{6} = 1\mathrm{ A.

Current through 3 ohm: I_3 = \frac{6}{3} = 2\mathrm{ A.

Power and Energy (OL/HL)

$$P = IV = I^2R = \frac{V^2}{R}$$ $$E = Pt = VIt$$

Example (OL): A 100 W light bulb is connected to a 230 V supply. Find the current and Resistance.

I = \frac{P}{V} = \frac{100}{230} \approx 0.435\mathrm{ A R = \frac{V}{I} = \frac{230}{0.435} \approx 529\mathrm{ ohms

Maximum Power Transfer (HL)

The maximum power is delivered to a load when the load resistance equals the internal resistance of The source: $R = r$.

Proof: $P = \frac{\varepsilon^2 R}{(R+r)^2}$. Setting $\frac{dP}{dR} = 0$ gives $R = r$.

Capacitors (HL)

Capacitance

$$C = \frac{Q}{V}$$

Unit: farad (F).

Energy Stored in a Capacitor

$$E = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}$$

Charging and Discharging

Charging: $Q = Q_0(1 - e^{-t/RC})$

Discharging: $Q = Q_0 e^{-t/RC}$

The time constant $\tau = RC$ is the time for the charge to reach $1 - e^{-1} \approx 63.2\%$ of its Final value during charging (or fall to $36.8\%$ during discharging).

Example (HL): A 100\mathrm{ \mu F capacitor is charged through a 50\mathrm{ k\Omega resistor. Find the time constant and the time for the capacitor to reach 95% of full charge.

\tau = RC = 50000 \times 100 \times 10^{-6} = 5\mathrm{ s

Q = 0.95 Q_0 \implies 1 - e^{-t/5} = 0.95 \implies e^{-t/5} = 0.05 \implies \frac{t}{5} = -\ln 0.05 \approx 3.0 \implies t \approx 15\mathrm{ s.

AC Circuits (HL)

Alternating Current

$$V = V_0 \sin(\omega t)$$

Where $V_0$ is the peak voltage and $\omega = 2\pi f$.

RMS Values

V_{\mathrm{rms} = \frac{V_0}{\sqrt{2}} I_{\mathrm{rms} = \frac{I_0}{\sqrt{2}}

The mains supply in Ireland is 230\mathrm{ V_{\mathrm{rms} at 50\mathrm{ Hz.

Power in AC Circuits

P_{\mathrm{avg} = V_{\mathrm{rms} I_{\mathrm{rms} \cos\phi

For purely resistive circuits: $\phi = 0$So P = V_{\mathrm{rms} I_{\mathrm{rms}.

Reactance and Impedance (HL)

Capacitive reactance:

$$X_C = \frac{1}{2\pi f C}$$

Inductive reactance:

$$X_L = 2\pi f L$$

Impedance of a series RLC circuit:

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

Resonance occurs when $X_L = X_C$:

$$F_0 = \frac{1}{2\pi\sqrt{LC}}$$

At resonance, impedance is minimum ($Z = R$) and current is maximum.

Example (HL): An RLC circuit has R = 100\mathrm{ ohms, L = 0.5\mathrm{ H C = 20\mathrm{ \mu F. Find the resonant frequency and the impedance at resonance.

F_0 = \frac{1}{2\pi\sqrt{0.5 \times 20 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-5}}} = \frac{1}{2\pi \times 3.162 \times 10^{-3}} \approx 50.3\mathrm{ Hz

At resonance: Z = R = 100\mathrm{ ohms.

Electromagnetic Induction (HL)

Faraday’s Law

The induced EMF is equal to the negative rate of change of magnetic flux:

\mathrm{EMF = -\frac{d\Phi}{dt}

Where $\Phi = BA\cos\theta$ is the magnetic flux.

Lenz’s Law

The direction of the induced current opposes the change in flux that produced it.

Transformers

For an ideal transformer with $N_p$ primary turns and $N_s$ secondary turns:

$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$ V_p I_p = V_s I_s \quad \mathrm{(energy conservation)

Common Pitfalls

1. Internal resistance — terminal voltage is less than EMF when current flows.
2. Series vs parallel — current is the same in series; voltage is the same in parallel.
3. Capacitors in series add like parallel resistors, and vice versa.
4. RMS vs peak — always use RMS values for power calculations with AC.
5. Lenz’s law — the minus sign in Faraday’s law represents this opposition.
6. Kirchhoff’s laws — apply the junction rule at every junction and the loop rule to every loop.

Practice Questions

Ordinary Level

1. A 9 V battery is connected to a 3 ohm resistor. Find the current and power.
2. A 4 ohm and a 12 ohm resistor are connected in parallel. Find the total resistance.
3. A kettle rated at 2000 W is used for 5 minutes. Find the energy transferred in joules.
4. Explain Ohm’s law and state when it does not apply.

Higher Level

1. A cell of EMF 6 V and internal resistance 0.5 ohms is connected to an external circuit of 11.5 ohms. Find the current and terminal voltage.

2. A 47\mathrm{ \mu F capacitor is charged to 12 V and then discharged through a 100\mathrm{ k\Omega resistor. Find the time constant and the charge after 10 s.

3. An AC circuit has V = 230\mathrm{ V_{\mathrm{rms} at 50\mathrm{ Hz connected to a 100\mathrm{ ohm resistor in series with a 0.1\mathrm{ H inductor. Find the impedance and the current.

4. A transformer has 500 primary turns and 50 secondary turns. The primary voltage is 230\mathrm{ V_{\mathrm{rms}. Find the secondary voltage. If the secondary current is 10 A, find the primary current.

5. Two capacitors of 22\mathrm{ \mu F and 47\mathrm{ \mu F are connected in series across a 12\mathrm{ V supply. Find the total capacitance and the charge on each capacitor.

6. An AC circuit has R = 200\mathrm{ ohms$$L = 0.2\mathrm{ H$$C = 50\mathrm{ \mu F at f = 100\mathrm{ Hz. Calculate X_L$$X_C$$ZThe phase angle, and the power dissipated.

7. A battery of EMF 12\mathrm{ V and internal resistance 1.0\mathrm{ \Omega is connected to an external circuit of resistance 5\mathrm{ \Omega. Calculate the power dissipated in the external circuit and the power dissipated in the internal resistance.

8. A step-up transformer converts 230\mathrm{ V to 11,500\mathrm{ V with 98% efficiency. Find the turns ratio and the primary current when the secondary current is 2\mathrm{ A.

9. Worked Example: Complex RC Circuit Analysis (HL)

A 100\mu\mathrm{F capacitor is charged through a 33\mathrm{ k\Omega resistor from a 9\mathrm{ V Supply.

(a) Time constant: \tau = RC = 33000 \times 100 \times 10^{-6} = 3.3\mathrm{ s

(b) Voltage after 1\mathrm{ s: V = 9(1 - e^{-1/3.3}) = 9(1 - 0.741) = 9 \times 0.259 = 2.33\mathrm{ V

(c) Current after 1\mathrm{ s: I = I_0 e^{-t/\tau} = \frac{9}{33000} \times e^{-1/3.3} = 2.727 \times 10^{-4} \times 0.741 = 2.02 \times 10^{-4}\mathrm{ A = 0.202\mathrm{ mA

(d) Time to reach 95\mathrm{ V across the capacitor: Already 9\mathrm{ VSo 95\mathrm{ V is Impossible. Assuming 8.55\mathrm{ V (95% of 9\mathrm{ V):

0.95 = 1 - e^{-t/3.3} \implies e^{-t/3.3} = 0.05 \implies t = -3.3 \times\ln(0.05) = -3.3 \times (-2.996) = 9.89\mathrm{ s

10. Worked Example: RLC Circuit Analysis (HL)

An RLC series circuit has R = 200\mathrm{ \Omega$$L = 0.5\mathrm{ H$$C = 20\mathrm{ \mu\mathrm{F At f = 100\mathrm{ Hz.

Inductive reactance: X_L = 2\pi fL = 2\pi \times 100 \times 0.5 = 314.2\mathrm{ \Omega

Capacitive reactance: X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 100 \times 20 \times 10^{-6}} = \frac{1}{0.01257} = 79.6\mathrm{ \Omega

Impedance: Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{200^2 + (314.2 - 79.6)^2} = \sqrt{40000 + 55030} = \sqrt{95030} = 308.3\mathrm{ \Omega

Phase angle: $\phi = \arctan\left(\frac{X_L - X_C}{R}\right) = \arctan\left(\frac{234.6}{200}\right) = \arctan(1.173) = 49.5^{\circ}$

The current lags the voltage by $49.5^{\circ}$.

Current: I = \frac{V}{Z} = \frac{230}{308.3} = 0.746\mathrm{ A

Power dissipated: P = I^2 R = 0.746^2 \times 200 = 111.3\mathrm{ W

11. Worked Example: Electromagnetic Induction (HL)

A coil of 200 turns and area 0.04\mathrm{ m^2 is in a magnetic field of flux density 0.5\mathrm{ T. The field is reduced to zero uniformly in 0.05\mathrm{ s.

\Delta\Phi = BA = 0.5 \times 0.04 = 0.02\mathrm{ Wb

\varepsilon = N\frac{\Delta\Phi}{\Delta t} = 200 \times \frac{0.02}{0.05} = 200 \times 0.4 = 80\mathrm{ V

By Lenz’s law, the induced EMF opposes the change in flux, so it acts to try to maintain the Original field.

12. Worked Example: Transformer Efficiency (HL)

A step-up transformer has 500 primary turns and 5000 secondary turns. The primary voltage is 230\mathrm{ V_{\mathrm{rms} and the secondary current is 5\mathrm{ A at 95\mathrm{ \Omega load.

Secondary voltage: V_s = 230 \times \frac{5000}{500} = 2300\mathrm{ V

Ideal secondary current: $I_s = \frac{V_p}{V_s} \times I_p$ (for ideal). Since efficiency is 95\mathrm{ \%:

$I_s \times V_s = 0.95 \times I_p \times V_p$

I_p = \frac{I_s \times V_s}{0.95 \times V_p} = \frac{5 \times 2300}{0.95 \times 230} = \frac{11500}{218.5} = 52.6\mathrm{ A

Power input: P_p = V_p I_p = 230 \times 52.6 = 12098\mathrm{ W \approx 12.1\mathrm{ kW

Power output: P_s = V_s I_s = 2300 \times 5 = 11500\mathrm{ W = 11.5\mathrm{ kW

Efficiency check: \frac{11500}{12098} = 0.950 = 95\mathrm{ \%. Correct.

13. Capacitors in Series and Parallel: Extended Analysis (HL)

Worked Example

Three capacitors C_1 = 22\mu\mathrm{F, C_2 = 47\mu\mathrm{FAnd C_3 = 10\mu\mathrm{F are Connected in series across a 12\mathrm{ V supply.

Total capacitance:

\frac{1}{C_{\mathrm{total}} = \frac{1}{22} + \frac{1}{47} + \frac{1}{10} = 0.0455 + 0.0213 + 0.1000 = 0.1668

C_{\mathrm{total} = \frac{1}{0.1668} = 5.996\mu\mathrm{F \approx 6.0\mu\mathrm{F

Charge (same on each): Q = C_{\mathrm{total}V = 5.996 \times 10^{-6} \times 12 = 7.195 \times 10^{-5}\mathrm{ C

Voltage across each:

V_1 = \frac{Q}{C_1} = \frac{71.95}{22} = 3.27\mathrm{ V

V_2 = \frac{Q}{C_2} = \frac{71.95}{47} = 1.53\mathrm{ V

V_3 = \frac{Q}{C_3} = \frac{71.95}{10} = 7.20\mathrm{ V

Check: 3.27 + 1.53 + 7.20 = 12.0\mathrm{ V. Correct.

Energy stored: E = \frac{1}{2}C_{\mathrm{total}V^2 = \frac{1}{2} \times 5.996 \times 10^{-6} \times 144 = 4.32 \times 10^{-4}\mathrm{ J

14. Kirchhoff’s Laws: Worked Example (HL)

A circuit has a 12\mathrm{ V battery (internal resistance 1\mathrm{ \Omega), a 6\mathrm{ \Omega Resistor in series, and a parallel combination of 8\mathrm{ \Omega and 12\mathrm{ \Omega Resistors.

Parallel resistance: $\frac{1}{R_p} = \frac{1}{8} + \frac{1}{12} = \frac{5}{24}$So R_p = 4.8\mathrm{ \Omega

Total external resistance: R = 6 + 4.8 = 10.8\mathrm{ \Omega

Total resistance including internal: R_{\mathrm{total} = 10.8 + 1 = 11.8\mathrm{ \Omega

Total current: I = \frac{12}{11.8} = 1.017\mathrm{ A

Terminal PD: V = 12 - 1.017 \times 1 = 10.98\mathrm{ V

Current through 8\mathrm{ \Omega: I_8 = \frac{10.98}{8} = 1.37\mathrm{ A

Current through 12\mathrm{ \Omega: I_{12} = \frac{10.98}{12} = 0.915\mathrm{ A

Check: 1.37 + 0.915 = 2.285\mathrm{ A. But the total current through the parallel combination Should equal the current through the 6\mathrm{ \Omega resistor, which is I = 1.017\mathrm{ A.

Wait — the current through the 6\mathrm{ \Omega resistor equals the total current from the Battery:

V_6 = 12 - 1.017 \times 1 - 1.017 \times 6 = 12 - 1.017 - 6.102 = 4.881\mathrm{ V

I_6 = \frac{4.881}{6} = 0.814\mathrm{ A

I_8 = \frac{4.881}{8} = 0.610\mathrm{ A

I_{12} = \frac{4.881}{12} = 0.407\mathrm{ A

Check: 0.814 + 0.610 + 0.407 = 1.831\mathrm{ A. But this should equal the total current 1.017\mathrm{ A.

There is an error — the terminal PD across the parallel combination should use the current through The 6\mathrm{ \Omega resistor, not the total current. Let me recalculate:

V_p = 1.017 \times 4.8 = 4.882\mathrm{ V

I_8 = \frac{4.882}{8} = 0.610\mathrm{ A, I_{12} = \frac{4.882}{12} = 0.407\mathrm{ A

Check: 0.610 + 0.407 = 1.017\mathrm{ A. Correct.

15. Summary Table: Key Electricity Formulas

Topic	Formula	Level	Notes
Current	$I = Q/t$	OL/HL	Rate of charge flow
Ohm’s law	$V = IR$	OL/HL	Ohmic conductors only
Resistivity	$R = \rho L/A$	OL/HL	Material property
Internal resistance	$V = \varepsilon - Ir$	HL	Terminal PD
Power	$P = IV = I^2R$	OL/HL	Three equivalent forms
Capacitance	$C = Q/V$	HL	Unit: farad
Capacitor energy	$E = \frac{1}{2}CV^2$	HL	Three equivalent forms
RC time constant	$\tau = RC$	HL	63.2% charge in one $\tau$
Reactance	$X_C = 1/(2\pi fC)$	HL	Decreases with frequency
Reactance	$X_L = 2\pi fL$	HL	Increases with frequency
Impedance	$Z = \sqrt{R^2 + (X_L-X_C)^2}$	HL	For RLC circuits
Resonance	$f_0 = 1/(2\pi\sqrt{LC})$	HL	$Z$ is minimum
Faraday’s law	$\varepsilon = -d\Phi/dt$	HL	Lenz’s law for direction
Transformer	$V_s/V_p = N_s/N_p$	HL	AC only

16. Practice Questions (Additional)

Higher Level (Additional)

11. A battery of EMF 9\mathrm{ V and internal resistance 1.5\mathrm{ \Omega is connected to a 4\mathrm{ \Omega resistor and a 12\mathrm{ \Omega resistor in parallel. Calculate the current through each resistor and the power dissipated in the 4\mathrm{ \Omega resistor.

12. A 47\mu\mathrm{F capacitor is charged to 20\mathrm{ V and then discharged through a 68\mathrm{ k\Omega resistor. Calculate (a) the time constant, (b) the voltage after 5\mathrm{ s(c) the energy stored initially, and (d) the energy remaining after 5\mathrm{ s.

13. An AC circuit has R = 150\mathrm{ \Omega$$L = 0.3\mathrm{ H$$C = 30\mathrm{ \mu\mathrm{F at f = 200\mathrm{ Hz. Calculate X_L$$X_C$$ZThe phase angle, the current, and the power dissipated.

14. A transformer with 1000 primary turns and 50 secondary turns is connected to a 230\mathrm{ V_{\mathrm{rms} mains supply. If the secondary delivers 10\mathrm{ A to a 100\mathrm{ \Omega load at 92\mathrm{ \% efficiency, calculate the primary current and the power wasted in the transformer.

15. Explain the difference between a capacitor and an inductor in an AC circuit. Include a description of how the impedance of each changes with frequency.

16. A potential divider circuit consists of a 20\mathrm{ k\Omega resistor and an NTC thermistor in series with a 12\mathrm{ V supply. The thermistor has resistance 15\mathrm{ k\Omega at 20^{\circ}\mathrm{C and 3\mathrm{ k\Omega at 80^{\circ}\mathrm{C. Calculate the output voltage at each temperature and explain how this circuit could be used as a temperature sensor.

17. Two capacitors of 33\mu\mathrm{F and 100\mu\mathrm{F are connected in series and then in parallel with a 6\mathrm{ V supply. For each configuration, find the total capacitance, the charge stored energy, and the PD across each capacitor.

18. Describe Faraday’s law of electromagnetic induction. Explain Lenz’s law and why it is a consequence of conservation of energy. Include a description of one experimental demonstration.

19. A coil of 150 turns, area 0.05\mathrm{ m^2Resistance 5\mathrm{ \OmegaIs in a magnetic field of flux density 0.6\mathrm{ T. The field is reversed in 0.02\mathrm{ s. Calculate the average EMF induced.

20. Explain why the national grid uses step-up and step-down transformers. Include a quantitative comparison of power losses at 11\mathrm{ kV and 400\mathrm{ kV for a cable of resistance 5\mathrm{ \Omega transmitting 500\mathrm{ MW.

Extended Worked Examples

Example 21: AC Circuit with Purely Resistive, Inductive, and Capacitive Loads

Compare the behaviour of a $100 \Omega$ resistor, a 0.5 \mathrm{ H inductor, and a 10 \mu\mathrm{F Capacitor when connected individually to a 230 \mathrm{ V$$50 \mathrm{ Hz AC supply.

Resistor:

I_R = \frac{V}{R} = \frac{230}{100} = 2.3 \mathrm{ A

Phase angle: $0^\circ$ (current and voltage in phase)

P_R = VI = 230 \times 2.3 = 529 \mathrm{ W

Inductor:

$X_L = 2\pi f L = 2\pi \times 50 \times 0.5 = 157.1 \Omega$

I_L = \frac{V}{X_L} = \frac{230}{157.1} = 1.464 \mathrm{ A

Phase angle: $90^\circ$ (current lags voltage)

P_L = 0 \mathrm{ W (pure inductor dissipates no power)

Capacitor:

$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 10 \times 10^{-6}} = \frac{1}{3.142 \times 10^{-3}} = 318.3 \Omega$

I_C = \frac{V}{X_C} = \frac{230}{318.3} = 0.723 \mathrm{ A

Phase angle: $-90^\circ$ (current leads voltage)

P_C = 0 \mathrm{ W (pure capacitor dissipates no power)

:::info Only resistors dissipate power in AC circuits. Inductors and capacitors store and release Energy each cycle but have zero average power dissipation. The power factor $\cos\phi$ accounts for This: a purely resistive load has $\cos\phi = 1$While purely reactive loads have $\cos\phi = 0$. :::

Example 22: Measuring Unknown Resistance with a Wheatstone Bridge

A Wheatstone bridge has P = 100 \Omega$$Q = 200 \Omega$$R = 150 \OmegaAnd $S$ is unknown. The galvanometer shows zero deflection. Calculate $S$. If the battery EMF is 6 \mathrm{ V Calculate the current through each resistor.

Step 1: Balance condition

$\frac{P}{Q} = \frac{R}{S} \implies \frac{100}{200} = \frac{150}{S} \implies S = 300 \Omega$

Step 2: Currents (no current through galvanometer at balance)

Branch 1 ($P$ and $Q$ in series): R_1 = 300 \Omega$$I_1 = 6/300 = 0.020 \mathrm{ A

Branch 2 ($R$ and $S$ in series): R_2 = 450 \Omega$$I_2 = 6/450 = 0.0133 \mathrm{ A

Total current from battery: I = I_1 + I_2 = 0.020 + 0.0133 = 0.0333 \mathrm{ A = 33.3 \mathrm{ mA

Step 3: Power in each resistor

P_P = I_1^2 \times P = 0.0004 \times 100 = 0.04 \mathrm{ W

P_Q = I_1^2 \times Q = 0.0004 \times 200 = 0.08 \mathrm{ W

P_R = I_2^2 \times R = 0.000177 \times 150 = 0.0266 \mathrm{ W

P_S = I_2^2 \times S = 0.000177 \times 300 = 0.0531 \mathrm{ W

Total power: P_{\mathrm{total} = 0.04 + 0.08 + 0.0266 + 0.0531 = 0.1997 \mathrm{ W \approx 0.2 \mathrm{ W

Check: P = VI = 6 \times 0.0333 = 0.1998 \mathrm{ W. Confirmed.

Example 23: Capacitor Charging Through a Resistor

A 47 \mu\mathrm{F capacitor is charged through a 100 \mathrm{ k\Omega resistor from a 12 \mathrm{ V supply. Calculate (a) the time constant, (b) the voltage after 5 \mathrm{ s(c) the Current after 5 \mathrm{ sAnd (d) the time to reach $95\%$ of the supply voltage.

Step 1: Time constant

\tau = RC = 100 \times 10^3 \times 47 \times 10^{-6} = 4.7 \mathrm{ s

Step 2: Voltage after 5 \mathrm{ s

$V(t) = V_0(1 - e^{-t/\tau}) = 12(1 - e^{-5/4.7}) = 12(1 - e^{-1.064})$

= 12(1 - 0.3452) = 12 \times 0.6548 = 7.86 \mathrm{ V

Step 3: Current after 5 \mathrm{ s

$I(t) = I_0 e^{-t/\tau} = \frac{V_0}{R} e^{-t/\tau} = \frac{12}{100000} \times e^{-1.064}$

= 0.00012 \times 0.3452 = 4.14 \times 10^{-5} \mathrm{ A = 41.4 \mu\mathrm{A

Step 4: Time to reach 95% of supply voltage

$0.95 = 1 - e^{-t/\tau} \implies e^{-t/\tau} = 0.05$

t = -\tau \ln(0.05) = -4.7 \times (-2.996) = 14.1 \mathrm{ s

This is approximately $3\tau$ (3 \times 4.7 = 14.1 \mathrm{ s).

:::info Useful rules of thumb for RC circuits:

• After $1\tau$: charged to $63.2\%$ of final value
• After $2\tau$: charged to $86.5\%$
• After $3\tau$: charged to $95.0\%$
• After $5\tau$: charged to $99.3\%$ (effectively fully charged) :::

Common Pitfalls Extended

Pitfall 6: Confusing Peak and RMS Values in AC Circuits

For a sinusoidal AC supply:

• Peak voltage: $V_0$
• RMS voltage: V_{\mathrm{rms} = V_0/\sqrt{2}
• Peak-to-peak voltage: $2V_0$

Mains electricity in Ireland is 230 \mathrm{ V RMS, which corresponds to a peak voltage of 230\sqrt{2} = 325 \mathrm{ V. Using the wrong value in calculations is a very common error.

Pitfall 7: Incorrectly Applying Kirchhoff’s Laws to AC Circuits

Kirchhoff’s laws apply to AC circuits, but voltages and currents must be added as phasors (taking Account of phase differences), not as simple algebraic sums. In a series RLC circuit: V_{\mathrm{total} \ne V_R + V_L + V_C; instead, V_{\mathrm{total}^2 = V_R^2 + (V_L - V_C)^2.

Pitfall 8: Forgetting That Capacitors Block DC

In a DC circuit at steady state, a fully charged capacitor draws no current and acts as an open Circuit. All the supply voltage appears across the capacitor. In an AC circuit, the capacitor Continuously charges and discharges, allowing AC current to flow. This is why capacitors are used to Block DC while passing AC signals.

Additional Practice Problems

21. A coil of inductance 0.2 \mathrm{ H and resistance $10 \Omega$ is connected to a 12 \mathrm{ V DC supply. Calculate (a) the initial rate of change of current, (b) the final steady-state current, and (c) the time constant of the circuit.

22. A 6 \mathrm{ V battery with internal resistance $0.5 \Omega$ is connected to two parallel resistors of $4 \Omega$ and $12 \Omega$. Calculate the current through each resistor, the terminal PD, and the power delivered to each resistor.

23. An AC generator produces an EMF given by $\varepsilon = 170\sin(314t)$ V. Calculate (a) the peak voltage, (b) the RMS voltage, (c) the frequency, and (d) the peak voltage across a $50 \Omega$ resistor connected to the generator.

24. A capacitor of 22 \mu\mathrm{F is charged to 50 \mathrm{ V and then connected across an uncharged 47 \mu\mathrm{F capacitor. Calculate (a) the common final voltage, (b) the charge on each capacitor, and (c) the energy lost in the process.

25. Describe an experiment to determine the relationship between the current through a filament lamp and the potential difference across it. Include a circuit diagram description and explain how you would use the results to show that the lamp is a non-ohmic conductor.