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Physics - Electricity (Static Electricity, Current, Circuits, Heating/Chemical effects)

Grade 8ICSE

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. Charging by friction involves the transfer of electrons from one body to another, resulting in one becoming positively charged and the other negatively charged.

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Electric Current (II): The rate of flow of charge through a cross-section of a conductor. It is measured in Amperes (AA). Conventionally, current flows from the positive terminal to the negative terminal of a cell.

Simple circuit showing a battery and an ammeter to measure current flow.
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Ohm's Law: At a constant temperature, the current (II) flowing through a conductor is directly proportional to the potential difference (VV) across its ends, expressed as V=I×RV = I \times R.

Circuit diagram to verify Ohm's Law using a voltmeter in parallel and ammeter in series.
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Chemical Effects: When electric current passes through a conducting solution (electrolyte), chemical reactions occur. This process is called electrolysis, which is used in electroplating.

Diagram of an electrolytic cell showing electrodes immersed in an electrolyte.

📐Formulae

Q=n×eQ = n \times e

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

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

V=I×RV = I \times R

Rs=R1+R2+R3+…R_s = R_1 + R_2 + R_3 + \dots

1Rp=1R1+1R2+1R3+…\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots

H=I2RtH = I^2Rt

P=V×IP = V \times I

💡Examples

Problem 1:

Calculate the current flowing through a conductor if a charge of 450 C450\ C passes through it in 22 minutes.

Solution:

Given: Charge Q=450 CQ = 450\ C, Time t=2 minutes=2×60=120 st = 2\text{ minutes} = 2 \times 60 = 120\ s. Using the formula I=QtI = \frac{Q}{t}, we get I=450120=3.75 AI = \frac{450}{120} = 3.75\ A.

Explanation:

To find the current, the time must be converted from minutes to the SI unit of seconds before applying the formula I=QtI = \frac{Q}{t}.

Problem 2:

Two resistors of 6 Ω6\ \Omega and 12 Ω12\ \Omega are connected in parallel. Calculate their equivalent resistance.

Solution:

Given: R1=6 ΩR_1 = 6\ \Omega and R2=12 ΩR_2 = 12\ \Omega. In parallel: 1Rp=1R1+1R2⇒1Rp=16+112=2+112=312=14\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \Rightarrow \frac{1}{R_p} = \frac{1}{6} + \frac{1}{12} = \frac{2+1}{12} = \frac{3}{12} = \frac{1}{4}. Therefore, Rp=4 ΩR_p = 4\ \Omega.

Explanation:

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

Problem 3:

An electric heater of resistance 50 Ω50\ \Omega draws a current of 5 A5\ A. Calculate the heat produced in 1010 seconds.

Solution:

Given: R=50 ΩR = 50\ \Omega, I=5 AI = 5\ A, t=10 st = 10\ s. Using Joule's Law: H=I2Rt=(5)2×50×10=25×50×10=12,500 JH = I^2Rt = (5)^2 \times 50 \times 10 = 25 \times 50 \times 10 = 12,500\ J.

Explanation:

The heat produced is directly proportional to the square of the current, the resistance, and the time for which the current flows.

Problem 4:

Determine the total resistance and the current flowing through a circuit where a 12 V12\ V battery is connected to three resistors of 2 Ω2\ \Omega, 3 Ω3\ \Omega, and 5 Ω5\ \Omega connected in series.

Circuit diagram showing three resistors connected in series with a 12V battery.

Solution:

Total resistance Rs=R1+R2+R3R_s = R_1 + R_2 + R_3 Rs=2 Ω+3 Ω+5 Ω=10 ΩR_s = 2\ \Omega + 3\ \Omega + 5\ \Omega = 10\ \Omega Using Ohm's Law: I=VRsI = \frac{V}{R_s} I=12 V10 Ω=1.2 AI = \frac{12\ V}{10\ \Omega} = 1.2\ A The current flowing through the circuit is 1.2 A1.2\ A.

Explanation:

In a series circuit, the total resistance is the sum of individual resistances. Since the same current flows through all components in series, we divide the total voltage by the equivalent resistance to find the current.

Problem 5:

An electric bulb is rated 100 W,250 V100\ W, 250\ V. Calculate the resistance of its filament and the current it draws when connected to a 250 V250\ V supply.

Circuit diagram showing a 100W bulb connected to a 250V power source.

Solution:

Given: P=100 WP = 100\ W, V=250 VV = 250\ V We know P=V×IP = V \times I, so I=PVI = \frac{P}{V} I=100250=0.4 AI = \frac{100}{250} = 0.4\ A Now, using V=I×RV = I \times R, R=VIR = \frac{V}{I} R=2500.4=625 ΩR = \frac{250}{0.4} = 625\ \Omega The resistance is 625 Ω625\ \Omega and current is 0.4 A0.4\ A.

Explanation:

The power rating allows us to find the current using the power formula. Once the current and voltage are known, the resistance of the filament can be calculated using Ohm's Law.