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Electricity and Magnetism - Electrical quantities (Current, voltage and resistance)

Grade 11A LevelPhysics

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

🔑Concepts

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Electric Current (II) is the rate of flow of electric charge (QQ). It is measured in Amperes (AA) using an ammeter connected in series. Charge is measured in Coulombs (CC).

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Conventional current flows from the positive terminal to the negative terminal, whereas electrons (the actual charge carriers in metals) flow from negative to positive.

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Potential Difference (VV), often called voltage, is the work done per unit charge in moving a charge between two points. It is measured in Volts (VV) using a voltmeter connected in parallel.

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Electromotive Force (e.m.f.) is the energy supplied by a source (like a battery) in driving unit charge around a complete circuit. Its unit is also the Volt (VV).

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Resistance (RR) is the property of a component that opposes the flow of electric current. It is defined as the ratio of potential difference to current (R=VIR = \frac{V}{I}) and measured in Ohms (Ω\Omega).

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Ohm's Law states that the current through a conductor is directly proportional to the potential difference across it, provided physical conditions like temperature remain constant (V∝IV ∝ I).

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The resistance of a wire is directly proportional to its length (LL) and inversely proportional to its cross-sectional area (AA), given by the relationship R∝LAR \propto \frac{L}{A}.

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In a series circuit, the total resistance is the sum of individual resistances: Rtotal=R1+R2+...R_{total} = R_1 + R_2 + .... The current is the same at all points.

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In a parallel circuit, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances: 1Rtotal=1R1+1R2+...\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + .... The voltage across each branch is the same.

📐Formulae

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

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

V=I×RV = I \times R

P=I×VP = I \times V

E=I×V×tE = I \times V \times t

Rseries=R1+R2+R3R_{series} = R_1 + R_2 + R_3

1Rparallel=1R1+1R2+1R3\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

R=ρLAR = \rho \frac{L}{A}

💡Examples

Problem 1:

A charge of 30C30 C passes through a light bulb in 11 minute. Calculate the current flowing through the bulb.

Solution:

I=Qt=30C60s=0.5AI = \frac{Q}{t} = \frac{30 C}{60 s} = 0.5 A

Explanation:

Current is the rate of flow of charge. Ensure the time is converted from minutes to seconds (11 min = 6060 s) to get the answer in Amperes.

Problem 2:

A resistor has a potential difference of 12V12 V across it and a current of 3A3 A flowing through it. What is its resistance?

Solution:

R=VI=12V3A=4ΩR = \frac{V}{I} = \frac{12 V}{3 A} = 4 \Omega

Explanation:

Using Ohm's Law, resistance is the ratio of potential difference to current.

Problem 3:

Two resistors, R1=6ΩR_1 = 6 \Omega and R2=3ΩR_2 = 3 \Omega, are connected in parallel. Calculate the total resistance of the circuit.

Solution:

1Rtotal=16+13=16+26=36  ⟹  Rtotal=63=2Ω\frac{1}{R_{total}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} \implies R_{total} = \frac{6}{3} = 2 \Omega

Explanation:

In a parallel circuit, the total resistance is always less than the smallest individual resistor. The reciprocal of the total resistance is the sum of the reciprocals of the individual resistances.