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The Particulate Nature of Matter - Current and Circuits

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electric current is the rate of flow of charge, where I=ΔQΔtI = \frac{\Delta Q}{\Delta t}. In metallic conductors, current is the drift of free electrons through a lattice of positive ions.

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Ohm's Law states that for a conductor at constant temperature, the current is directly proportional to the potential difference across it, expressed as V=IRV = IR.

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Resistivity (ρ\rho) is an intrinsic property of a material. Resistance (RR) depends on the material's resistivity, its length (LL), and its cross-sectional area (AA) as defined by R=ρLAR = \rho \frac{L}{A}.

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Kirchhoff's First Law (Current Law) is a consequence of the conservation of charge: the sum of currents entering a junction equals the sum of currents leaving the junction (∑I=0\sum I = 0).

Junction diagram illustrating Kirchhoff's Current Law where I1 = I2 + I3.
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Kirchhoff's Second Law (Voltage Law) is a consequence of the conservation of energy: the algebraic sum of the electromotive forces (EMF) in any closed loop is equal to the algebraic sum of the potential drops (∑ϵ=∑IR\sum \epsilon = \sum IR).

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Potential dividers use two or more resistors in series to provide a fraction of the source voltage. The output voltage VoutV_{out} is determined by the ratio of the resistances.

Circuit diagram of a potential divider network.

📐Formulae

I=ΔQΔtI = \frac{\Delta Q}{\Delta t}

I=nAvqI = nAvq

V=WqV = \frac{W}{q}

R=VIR = \frac{V}{I}

R=ρLAR = \rho \frac{L}{A}

P=VI=I2R=V2RP = VI = I^2R = \frac{V^2}{R}

ϵ=I(R+r)\epsilon = I(R + r)

Rseries=R1+R2+...R_{series} = R_1 + R_2 + ...

1Rparallel=1R1+1R2+...\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + ...

💡Examples

Problem 1:

A copper wire has a cross-sectional area of 2.0×10−6m22.0 \times 10^{-6} m^2 and carries a current of 5.0A5.0 A. If the number density of free electrons in copper is 8.5×1028m−38.5 \times 10^{28} m^{-3}, calculate the drift velocity of the electrons.

Solution:

Using the formula I=nAvqI = nAvq, we rearrange for vv: v=InAqv = \frac{I}{nAq} v=5.0(8.5×1028)(2.0×10−6)(1.6×10−19)v = \frac{5.0}{(8.5 \times 10^{28})(2.0 \times 10^{-6})(1.6 \times 10^{-19})} v≈1.84×10−4ms−1v \approx 1.84 \times 10^{-4} m s^{-1}

Explanation:

This demonstrates that while the signal of electricity travels near the speed of light, the actual 'particles' (electrons) move very slowly through the lattice of the conductor.

Problem 2:

A battery with an EMF of 12.0V12.0 V and an internal resistance of 0.5Ω0.5 \Omega is connected to a resistor of 5.5Ω5.5 \Omega. Determine the terminal potential difference of the battery.

Solution:

First, find the total current II using ϵ=I(R+r)\epsilon = I(R + r): I=ϵR+r=12.05.5+0.5=2.0AI = \frac{\epsilon}{R + r} = \frac{12.0}{5.5 + 0.5} = 2.0 A Now, find the terminal potential difference VV: V=ϵ−Ir=12.0−(2.0×0.5)=11.0VV = \epsilon - Ir = 12.0 - (2.0 \times 0.5) = 11.0 V

Explanation:

The terminal potential difference is lower than the EMF because some energy is dissipated as heat within the battery's internal resistance (IrIr 'lost volts').

Problem 3:

Calculate the power dissipated in a 10Ω10 \Omega resistor when it is connected in parallel with a 15Ω15 \Omega resistor, both of which are connected to a 30V30 V ideal power supply.

Parallel circuit diagram with a 30V source and two resistors of 10 Ohms and 15 Ohms.

Solution:

In a parallel circuit, the potential difference across each branch is equal to the supply voltage. Therefore, the voltage across the 10Ω10 \Omega resistor is: V=30VV = 30 V Using the power formula: P=V2RP = \frac{V^2}{R} P=30210P = \frac{30^2}{10} P=90010P = \frac{900}{10} P=90WP = 90 W

Explanation:

Because the resistors are in parallel and the supply is ideal (no internal resistance), each resistor experiences the full 30V30 V. The power dissipated depends only on that voltage and its individual resistance.

Problem 4:

A circuit consists of a 24.0V24.0 V ideal DC power supply connected in series with a 4.0Ω4.0 \Omega resistor (R1R_1) and a parallel combination of two resistors: a 12.0Ω12.0 \Omega resistor (R2R_2) and a 6.0Ω6.0 \Omega resistor (R3R_3). Calculate the total current flowing from the power supply and the potential difference across the parallel combination.

A circuit diagram showing a 24V source in series with a 4 ohm resistor, which is then in series with a parallel network of 12 ohm and 6 ohm resistors.

Solution:

Rp=(1R2+1R3)−1R_p = \left( \frac{1}{R_2} + \frac{1}{R_3} \right)^{-1} Rp=(112.0+16.0)−1=4.0ΩR_p = \left( \frac{1}{12.0} + \frac{1}{6.0} \right)^{-1} = 4.0 \Omega Rtotal=R1+Rp=4.0+4.0=8.0ΩR_{total} = R_1 + R_p = 4.0 + 4.0 = 8.0 \Omega Itotal=VRtotal=24.08.0=3.0AI_{total} = \frac{V}{R_{total}} = \frac{24.0}{8.0} = 3.0 A Vp=Itotal×Rp=3.0×4.0=12.0VV_p = I_{total} \times R_p = 3.0 \times 4.0 = 12.0 V

Explanation:

First, the equivalent resistance of the parallel branch (R2R_2 and R3R_3) is calculated using the reciprocal formula. This equivalent resistance is then added to the series resistor R1R_1 to find the total circuit resistance. Ohm's law is applied to the entire circuit to find the total current. Finally, the potential difference across the parallel section is found by multiplying the total current by the equivalent resistance of that specific section.