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

Grade 11IGCSEPhysics

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

πŸ”‘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 (Ξ©\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.

πŸ“Formulae

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

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

V=IΓ—RV = I \times R

P=IΓ—VP = I \times V

E=IΓ—VΓ—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}

πŸ’‘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=4Ξ©R = \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=6Ξ©R_1 = 6 \Omega and R2=3Ξ©R_2 = 3 \Omega, are connected in parallel. Calculate the total resistance of the circuit.

Solution:

1Rtotal=16+13=16+26=36β€…β€ŠβŸΉβ€…β€ŠRtotal=63=2Ξ©\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.

Electrical quantities (Current, voltage and resistance) Revision - Grade 11 Physics IGCSE