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Electricity - Electric current and potential difference

Grade 10CBSE

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

🔑Concepts

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Electric Current (II): The rate of flow of electric charges through a conductor. It is defined as I=QtI = \frac{Q}{t}, where QQ is net charge and tt is time.

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The SI unit of electric current is Ampere (AA). One Ampere is the flow of one Coulomb of charge per second: 1A=1C/1s1 A = 1 C / 1 s.

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Electric Charge (QQ): The property of matter that causes it to experience a force in an electromagnetic field. Its SI unit is Coulomb (CC).

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Quantization of Charge: The total charge is an integral multiple of the charge of an electron, represented by Q=neQ = ne, where e≈1.6×10−19Ce \approx 1.6 \times 10^{-19} C.

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Electric Potential Difference (VV): The work done (WW) to move a unit charge from one point to another in an electric circuit. V=WQV = \frac{W}{Q}.

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The SI unit of potential difference is Volt (VV). One Volt is defined as 11 Joule of work done to move 11 Coulomb of charge: 1V=1J/1C1 V = 1 J / 1 C.

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Ammeter: A device used to measure electric current; it has low resistance and is always connected in series with the circuit.

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Voltmeter: A device used to measure potential difference; it has high resistance and is always connected in parallel across the points where the potential difference is to be measured.

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Direction of Current: Conventionally, the direction of electric current is taken as the direction of flow of positive charges (from positive terminal to negative terminal), which is opposite to the direction of the flow of electrons.

📐Formulae

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

Q=n⋅eQ = n \cdot e

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

1 A=1 C1 s1 \text{ A} = \frac{1 \text{ C}}{1 \text{ s}}

1 V=1 J1 C1 \text{ V} = \frac{1 \text{ J}}{1 \text{ C}}

💡Examples

Problem 1:

A current of 0.5 A0.5 \text{ A} is drawn by a filament of an electric bulb for 1010 minutes. Find the amount of electric charge that flows through the circuit.

Solution:

Given: I=0.5 AI = 0.5 \text{ A}, t=10 minutes=10×60=600 st = 10 \text{ minutes} = 10 \times 60 = 600 \text{ s}. Using Q=I×tQ = I \times t: Q=0.5 A×600 s=300 CQ = 0.5 \text{ A} \times 600 \text{ s} = 300 \text{ C}.

Explanation:

To find the charge, we multiply the current by the time in seconds. 1010 minutes must be converted to 600600 seconds for standard SI units.

Problem 2:

How much work is done in moving a charge of 2 C2 \text{ C} across two points having a potential difference of 12 V12 \text{ V}?

Solution:

Given: Q=2 CQ = 2 \text{ C}, V=12 VV = 12 \text{ V}. Using W=V×QW = V \times Q: W=12 V×2 C=24 JW = 12 \text{ V} \times 2 \text{ C} = 24 \text{ J}.

Explanation:

Work done is the product of the potential difference and the amount of charge moved.

Problem 3:

Calculate the number of electrons constituting one Coulomb of charge.

Solution:

Given: Q=1 CQ = 1 \text{ C}, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}. Using n=Qen = \frac{Q}{e}: n=11.6×10−19≈6.25×1018n = \frac{1}{1.6 \times 10^{-19}} \approx 6.25 \times 10^{18} electrons.

Explanation:

By applying the quantization of charge formula Q=neQ = ne, we find the total number of elementary charges required to sum up to one Coulomb.