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Electricity - ELECTRIC CURRENT AND CIRCUIT11.1 ELECTRIC CURRENT AND CIRCUIT11.1 ELECTRIC CURRENT AND CIRCUIT11.1 ELECTRIC CURRENT AND CIRCUIT11.1 ELECTRIC CURRENT AND CIRCUIT

Grade 10CBSE

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

🔑Concepts

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An electric circuit is a continuous and closed path of an electric current. If the path is broken anywhere (or the switch is turned off), the current stops flowing and the device (like a bulb) does not work.

A basic schematic diagram of an electric circuit comprising a cell, an electric bulb, an ammeter and a plug key.
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Electric current is expressed by the amount of charge flowing through a particular area in unit time. Conventionally, the direction of electric current is taken as opposite to the direction of the flow of electrons.

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The SI unit of electric charge is Coulomb (CC), which is equivalent to the charge contained in nearly 6×10186 \times 10^{18} electrons. The SI unit of current is Ampere (AA), named after Andre-Marie Ampere.

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An instrument called ammeter measures electric current in a circuit. It is always connected in series in a circuit through which the current is to be measured.

Circuit showing an ammeter connected in series with a resistor and a battery.

📐Formulae

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

Q=n×eQ = n \times e

1A=1C1s1 A = \frac{1 C}{1 s}

e=1.6×10−19Ce = 1.6 \times 10^{-19} C

💡Examples

Problem 1:

A current of 0.5A0.5 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: Current I=0.5AI = 0.5 A, Time t=10 mint = 10 \text{ min}. First, convert time to seconds: t=10×60=600st = 10 \times 60 = 600 s Using the formula: Q=I×tQ = I \times t Q=0.5A×600s=300CQ = 0.5 A \times 600 s = 300 C

Explanation:

To find the charge, we multiply the current by the time in seconds. The resulting unit is Coulombs (CC).

Problem 2:

Calculate the number of electrons constituting one coulomb of charge.

Solution:

Given: Total charge Q=1CQ = 1 C, Charge of one electron e=1.6×10−19Ce = 1.6 \times 10^{-19} C. Using the formula Q=neQ = ne, we find nn: n=Qen = \frac{Q}{e} n=1C1.6×10−19Cn = \frac{1 C}{1.6 \times 10^{-19} C} n=6.25×1018n = 6.25 \times 10^{18}

Explanation:

By dividing the total charge by the charge of a single electron, we determine that approximately 6.25×10186.25 \times 10^{18} electrons are required to make 1C1 C of charge.

Problem 3:

A steady current of 2A2 A flows through a circuit for 3030 seconds. If an ammeter is placed in the circuit as shown, determine the total charge QQ that passes through the ammeter during this interval.

Circuit diagram showing a $2 A$ current measured by an ammeter.

Solution:

Given: I=2AI = 2 A t=30st = 30 s

Using the formula Q=I×tQ = I \times t: Q=2A×30sQ = 2 A \times 30 s Q=60CQ = 60 C

The total charge flowing through the circuit is 60C60 C.

Explanation:

The electric charge is the product of the magnitude of the current and the time for which it flows through the conductor.

Problem 4:

An electric iron draws a current of 3.4A3.4 A from a 220V220 V source. If the iron is used for 22 hours, calculate the total electric charge QQ that flows through the heating element. Also, draw the circuit diagram representing this setup with an ammeter and a switch.

A circuit diagram showing a 220V battery source, an ammeter, a heating element (resistor), and a closed switch connected in a series loop.

Solution:

Given: Current I=3.4AI = 3.4 A Time t=2t = 2 hours =2×60×60s=7200s= 2 \times 60 \times 60 s = 7200 s

Using the formula for electric charge: Q=I×tQ = I \times t Q=3.4A×7200sQ = 3.4 A \times 7200 s Q=24480CQ = 24480 C

The total charge flowing through the element is 2448024480 Coulombs.

Explanation:

To find the total charge, we first convert the time from hours into the SI unit (seconds). By multiplying the steady current by the total time duration, we obtain the total quantity of charge that has passed through the cross-section of the conductor.