Topic 3 - Electric Circuits

Define potential difference

The energy transferred (or work done) per unit charge between two points in a circuit (V = W/Q)

Define emf

The work done per unit charge by a power supply, converting other forms of energy into electrical potential energy

Define current

The rate of flow of charge (I = ∆Q/∆t)

Derive I = Ne/∆t

Total charge ∆Q = Ne, where N is the number of electrons and e is the electron charge. Since I = ∆Q/∆t, substituting gives I = Ne/∆t

Calculate the number of electrons in a charge

Number of electrons, N = Total Charge (Q) / Charge on one electron (e) | N = Q / (1.6 × 10⁻¹⁹)

Define resistance

The ratio of potential difference across a component to the current flowing through it (R = V/I)

Derive the current ratio for parallel branches

In parallel, V is constant. V = I₁R₁ = I₂R₂. Therefore, I₁/I₂ = R₂/R₁. The current ratio is the inverse of the resistance ratio.

Derive the formula for resistors in series

Vtotal = V₁ + V₂ + V₃. Using V=IR and Iconstant: IRtotal = IR₁ + IR₂ + IR₃. Thus, Rtotal = R₁ + R₂ + R₃

Derive the formula for resistors in parallel

Itotal = I₁ + I₂ + I₃. Using I=V/R and Vconstant: V/Rtotal = V/R₁ + V/R₂ + V/R₃. Thus, 1/Rtotal = 1/R₁ + 1/R₂ + 1/R₃

State Kirchhoff's potential difference law

The sum of the emfs in any closed loop equals the sum of the potential differences. This is due to the conservation of energy.

State Kirchhoff's current law

The sum of currents entering a junction equals the sum of currents leaving it. This is due to the conservation of charge.

Define lost volts

The energy per unit charge dissipated within a power supply due to its internal resistance (Lost Volts = I × r)

Define terminal potential difference

The potential difference across the terminals of a power supply when current is flowing: V = EMF - Ir

Explain why an ideal voltmeter measures EMF directly

An ideal voltmeter has infinite resistance, so no current flows. Therefore, lost volts (Ir) is zero, and the terminal pd (V) equals the EMF.

Explain why a battery heats up with high current

A high current causes significant energy loss to the internal resistance (high lost volts, Ir). This energy is transferred as heat, increasing the battery's temperature.

Draw a circuit to find EMF and internal resistance

Circuit: A cell in series with an ammeter and a variable resistor. A voltmeter is connected in parallel across the cell (or across the variable resistor).

Describe the experiment to find EMF and internal resistance

Set up the circuit. Vary the variable resistor to get different current (I) and terminal pd (V) readings. Plot V (y-axis) against I (x-axis). The y-intercept is the EMF, the gradient is -r (so r = -gradient).

Explain the intercepts on the V-I graph for a cell

y-intercept (I=0): Terminal pd equals EMF (no lost volts). x-intercept (V=0): Short-circuit condition; all EMF is lost as heat in the internal resistance.

Define power

The rate of energy transfer (P = E/t)

Derive P = I²R and P = V²/R from P = VI

From P=VI and V=IR: P = I(IR) = I²R. From P=VI and I=V/R: P = V(V/R) = V²/R.

(ii) Explain bulb brightness after adding a parallel bulb, assuming internal resistance pf the battery

Brightness depends on power (P=V²/R_individual). Since the terminal pd (V) across AB decreases for all bulbs, the power and thus brightness of each bulb decreases.

Define the terms in I = nqvA

I: Current | n: Number density of charge carriers | q: Charge on each charge carrier | v: Drift velocity of charge carriers | A: Cross-sectional area of the conductor

Derive I = nqvA

Charge in a length l: ∆Q = (number of charge carriers in volume A×l) × q = (n A l) q. Time for this charge to pass: ∆t = l/v. I = ∆Q/∆t = (n A l q) / (l/v) = n A q v

Draw a circuit to find the resistance of a wire

Circuit A: Wire in series with a power supply and an ammeter. A voltmeter is connected in parallel across the wire. (A variable resistor can be added in series to vary current).

Determine resistance from a V-I graph

Resistance (R) at any point is the ratio V/I for that point. It is NOT the gradient of the line (which is 1/R for an ohmic conductor).

Draw I-V and V-I graphs for components

I-V: Ohmic (straight line), Filament bulb (curve flattening), Thermistor (steeper curve), Diode (flat on reverse, sharp rise after ~0.7V forward). V-I: Ohmic (straight line), Filament bulb (curve steepening), Thermistor (less steep curve), Diode (flat on reverse, sharp rise after threshold).

Explain the I-V relationship for a filament bulb

As current/pd increases, the wire heats up. Ions vibrate more, increasing collision frequency with electrons. This limits the increase in drift velocity (v), so current increases less than pd. Thus, R (=V/I) increases.

Explain thermistor resistance vs. temperature (external heating)

Increased temperature provides energy, promoting more electrons into the conduction band. This increases the number density (n) of charge carriers, decreasing resistance.

Explain the I-V relationship for a thermistor (self-heating)

Increased current causes self-heating. This increases the temperature, promoting more electrons into the conduction band (n increases). This causes a large increase in current, so R (=V/I) decreases.

Explain LDR resistance vs. light intensity

Higher light intensity means more photons per second. More electrons absorb energy and are promoted to the conduction band, increasing n. This increases current, thus decreasing resistance.

Explain the I-V graph of a diode

Low resistance in forward bias above a threshold voltage (~0.7V for Si). Very high resistance in reverse bias, until a breakdown voltage is reached.

Derive the potential divider formula

For two resistors R₁ and R₂ in series: Vout = Vin × [R₂ / (R₁ + R₂)]

Describe and draw a potential divider circuit

A circuit that splits a fixed input voltage across two resistors. Can use two fixed resistors or a single variable resistor (rheostat/potentiometer).

Explain the voltmeter reading as the slider moves from bottom to top (in a potentiometer circuit)

At the bottom: Vout = 0V. At the top: Vout = V_supply. The output voltage is proportional to the length of the resistive wire between the slider and the bottom.

Explain why pd is proportional to length in a potentiometer

For a uniform wire, R ∝ L. The current is constant. Since V = IR, it follows that V ∝ R ∝ l.

Define resistivity

The resistance of a 1m length of material with a 1m² cross-sectional area (R = ρl/A)

What does resistivity depend on?

It is a material property, independent of the object's dimensions. It depends on the material and its temperature.

How does a thermistor's resistivity change with temperature?

Resistivity decreases as temperature increases (due to a large increase in charge carrier density, n).

How does a metal's resistivity change with temperature?

Resistivity increases as temperature increases (due to increased lattice ion vibrations, reducing electron drift velocity, v).

Describe an experiment to find the resistivity of a wire

Measure R for different lengths (l) of wire. Measure diameter (d) with a micrometer to find A=π(d/2)². Plot R vs l. Gradient = ρ/A, so ρ = gradient × A.

Why in the cell practical for determining EMF and internal resistance should the voltmeter be ideal, but the ammeter not being ideal doesn’t affect the results

Ammeter explanation: If ammeter has resistance, current decreased but doesn’t affect the determination because current through cell/r is measured Or doesn’t affect the determination because the voltmeter measures the terminal p.d. for that current OR The resistance of the ammeter contributes to the load/circuit/total resistance Values of p.d. corresponding to given values of current will be unchanged Voltmeter explanation: If voltmeter has smaller resistance it would draw current measured current not current through cell/r