Electricity

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

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$$X_C$$Z The 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 $9\mathrm{ V}$ So $95\mathrm{ V}$ 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 $C_1 = 22\mu\mathrm{F}$, $C_2 = 47\mu\mathrm{F}$ And $C_3 = 10\mu\mathrm{F}$ 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

16. Practice Questions (Additional)

Higher Level (Additional)

1. 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.

2. 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 $5\mathrm{ s}$.

3. 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$$X_C$$Z The phase angle, the current, and the power dissipated.

4. 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.

5. 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.

6. 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.

7. 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.

8. 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.

9. A coil of 150 turns, area $0.05\mathrm{ m^2}$ Resistance $5\mathrm{ \Omega}$ Is 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.

10. 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)

Example 22: Measuring Unknown Resistance with a Wheatstone Bridge

A Wheatstone bridge has P = 100 \Omega$$Q = 200 \Omega$$R = 150 \Omega And $S$ 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 $5 \mathrm{ s}$ And (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 $3\tau$ ($3 \times 4.7 = 14.1 \mathrm{ s}$).

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

1. 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.

2. 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.

3. 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.

4. 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.

5. 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.

Cross-References

• Mechanics: Covers work, energy, and power concepts that connect to electrical energy and circuits.
• Light and Waves: Explores the electromagnetic spectrum and wave properties, linking to electromagnetic induction.
• Heat and Thermodynamics: Covers energy transfer and thermodynamics, connecting to the energy dissipation in electrical circuits.
• Practice Physics: Interactive practice problems covering electricity, circuits, and other physics topics.

Intuition

Electricity is the physics of charge in motion. The three fundamental quantities, voltage (energy per unit charge), current (rate of charge flow), and resistance (opposition to flow), are related by Ohm’s law: $V = IR$. Think of a circuit as a water system: voltage is the pump pressure, current is the flow rate, and resistance is the pipe narrowness. This analogy extends logically: resistors in series are like pipes in a chain (resistances add), while resistors in parallel are like parallel pipes (total resistance decreases).

Capacitors store energy in electric fields and behave very differently from resistors. A capacitor resists changes in voltage, it charges exponentially when connected to a supply and discharges exponentially when the supply is removed. The time constant $\tau = RC$ determines how fast this happens: after one $\tau$, a capacitor reaches 63.2% of its final charge; after $5\tau$, it is effectively fully charged. This exponential behaviour is why capacitors are used for timing circuits, smoothing power supply output, and filtering signals.

Electromagnetic induction is the principle behind generators and transformers. Faraday’s law tells you that a changing magnetic flux through a coil induces an EMF, and Lenz’s law tells you the induced current always opposes the change that caused it. Transformers exploit this to step voltage up or down: stepping up to high voltage for efficient long-distance transmission (reducing $I^2R$ losses), then stepping down to safe levels for domestic use. The relationship $V_s/V_p = N_s/N_p$ connects the voltage ratio to the turns ratio, and energy conservation requires $V_p I_p = V_s I_s$ for an ideal transformer.