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Physics - Electricity and Magnetism (Static Electricity, Current, Simple Circuits, Magnetic Field)

Grade 9ICSE

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

🔑Concepts

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Static Electricity: It is the study of electric charges at rest. According to the law of electrostatics, like charges repel each other, while unlike charges attract each other.

Diagram showing two positive charges repelling each other.
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Electric Current: The rate of flow of charge through a cross-section of a conductor. By convention, the direction of current is taken from the positive terminal to the negative terminal of the cell.

Simple circuit showing conventional current flow from positive to negative terminal.
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Potential Difference: The work done in moving a unit positive charge from one point to another in an electric field. It is measured in Volts (VV).

Diagram representing potential difference between two points A and B.
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Magnetic Field Lines: Imaginary lines around a magnet along which a North pole would move. They emerge from the North pole and enter the South pole externally.

Magnetic field lines emerging from the North pole and entering the South pole of a bar magnet.

📐Formulae

Q=neQ = ne

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

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

V=IRV = IR

R=ρlAR = \rho \frac{l}{A}

💡Examples

Problem 1:

Calculate the current flowing through a wire if a charge of 120 C120 \text{ C} flows through it in 22 minutes.

Solution:

Given: Q=120 CQ = 120 \text{ C}, t=2 minutes=2×60=120 st = 2 \text{ minutes} = 2 \times 60 = 120 \text{ s}. Using the formula I=QtI = \frac{Q}{t}, we get I=120120=1 AI = \frac{120}{120} = 1 \text{ A}.

Explanation:

To find the current, the time must be converted into the SI unit (seconds) before dividing the total charge by the time duration.

Problem 2:

How much work is done in moving a charge of 5 C5 \text{ C} across two points having a potential difference of 12 V12 \text{ V}?

Solution:

Given: Q=5 CQ = 5 \text{ C}, V=12 VV = 12 \text{ V}. Using the formula V=WQV = \frac{W}{Q}, we rearrange to find W=V×QW = V \times Q. Thus, W=12×5=60 JW = 12 \times 5 = 60 \text{ J}.

Explanation:

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

Problem 3:

A conductor has a resistance of 10 Ω10 \, \Omega. If a potential difference of 5 V5 \text{ V} is applied across its ends, calculate the current.

Solution:

Given: R=10 ΩR = 10 \, \Omega, V=5 VV = 5 \text{ V}. From Ohm's Law, I=VRI = \frac{V}{R}. So, I=510=0.5 AI = \frac{5}{10} = 0.5 \text{ A}.

Explanation:

Current is directly proportional to potential difference and inversely proportional to resistance as per V=IRV = IR.

Problem 4:

Determine the equivalent resistance of the circuit shown below where two resistors R1=4 ΩR_1 = 4\,\Omega and R2=6 ΩR_2 = 6\,\Omega are connected in series to a 10 V10\text{ V} battery.

Circuit diagram with a 10V source and two resistors of 4 ohms and 6 ohms in series.

Solution:

In a series circuit, the total resistance RsR_s is the sum of individual resistances: Rs=R1+R2R_s = R_1 + R_2 Rs=4 Ω+6 Ω=10 ΩR_s = 4\,\Omega + 6\,\Omega = 10\,\Omega The total current II is: I=VRs=10 V10 Ω=1 AI = \frac{V}{R_s} = \frac{10\text{ V}}{10\,\Omega} = 1\text{ A}

Explanation:

Since the resistors are connected end-to-end, the same current flows through both, and their resistances add up directly.

Problem 5:

Calculate the total current II flowing from the battery in the following parallel circuit where R1=10 ΩR_1 = 10\,\Omega and R2=10 ΩR_2 = 10\,\Omega are connected to a 5 V5\text{ V} source.

Circuit diagram with a 5V source and two 10 ohm resistors connected in parallel.

Solution:

For resistors in parallel, the equivalent resistance RpR_p is: 1Rp=1R1+1R2\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} 1Rp=110+110=210\frac{1}{R_p} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} Rp=5 ΩR_p = 5\,\Omega Total current II: I=VRp=5 V5 Ω=1 AI = \frac{V}{R_p} = \frac{5\text{ V}}{5\,\Omega} = 1\text{ A}

Explanation:

In a parallel circuit, the potential difference across each resistor is the same, but the total current is the sum of currents through each branch.