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Electricity: Magnetic and Heating Effects - Voltaic cell

Grade 8CBSE

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

🔑Concepts

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A simple Voltaic cell, also known as a Galvanic cell, converts chemical energy into electrical energy through redox reactions occurring between two different metals in an electrolyte.

A simple voltaic cell consisting of a Zinc anode and a Copper cathode immersed in dilute sulfuric acid.
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The electrode with higher reactivity (usually Zinc) acts as the negative terminal (Anode) where oxidation occurs: Zn→Zn2++2e−Zn \rightarrow Zn^{2+} + 2e^-.

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The electrode with lower reactivity (usually Copper) acts as the positive terminal (Cathode). Hydrogen ions (H+H^+) from the electrolyte move towards it to gain electrons: 2H++2e−→H2(g)2H^+ + 2e^- \rightarrow H_2(g).

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Electric current flows in the external circuit from the Copper electrode (positive) to the Zinc electrode (negative), while electrons flow from Zinc to Copper.

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The chemical action eventually creates local action and polarization, which are defects that reduce the efficiency of the cell over time.

📐Formulae

H=I2RtH = I^2 R t

V=I×RV = I \times R

P=V×IP = V \times I

1 Joule=1 Ampere2×1 Ohm×1 Second1 \text{ Joule} = 1 \text{ Ampere}^2 \times 1 \text{ Ohm} \times 1 \text{ Second}

💡Examples

Problem 1:

Calculate the heat produced in a wire of resistance 10Ω10 \Omega when a current of 2A2 A flows through it for 55 seconds.

Solution:

H=I2RtH = I^2 R t H=(2)2×10×5H = (2)^2 \times 10 \times 5 H=4×10×5=200 JH = 4 \times 10 \times 5 = 200 \text{ J}

Explanation:

We use Joule's Law of Heating. By substituting the given values of current (I=2AI = 2 A), resistance (R=10ΩR = 10 \Omega), and time (t=5st = 5 s), we find the total heat energy produced in Joules.

Problem 2:

In a simple Voltaic cell, if the potential difference (VV) is 1.08V1.08 V and the current (II) flowing through an external bulb is 0.2A0.2 A, what is the resistance (RR) of the bulb?

Solution:

R=VIR = \frac{V}{I} R=1.080.2R = \frac{1.08}{0.2} R=5.4ΩR = 5.4 \Omega

Explanation:

Using Ohm's Law, we rearrange the formula to solve for resistance by dividing the voltage by the current.

Problem 3:

Determine the total charge flow if a Voltaic cell provides a current of 0.5A0.5 A for 120120 seconds.

Solution:

Q=I×tQ = I \times t Q=0.5×120=60 CQ = 0.5 \times 120 = 60 \text{ C}

Explanation:

Electric charge (QQ) is the product of the electric current (II) and the time (tt) for which it flows. The unit is Coulombs (CC).

Problem 4:

A student sets up a voltaic cell using Zinc and Copper plates. If the cell produces a constant current of I=0.3AI = 0.3 A to light a small LED for t=200st = 200 s, calculate the total heat energy dissipated in the circuit if the internal resistance of the setup is R=4ΩR = 4 \Omega.

A voltaic cell connected to a resistive circuit generating heat.

Solution:

Given: Current (II) = 0.3A0.3 A Time (tt) = 200s200 s Resistance (RR) = 4Ω4 \Omega

Using the formula for heating effect: H=I2RtH = I^2 R t H=(0.3)2×4×200H = (0.3)^2 \times 4 \times 200 H=0.09×4×200H = 0.09 \times 4 \times 200 H=0.36×200H = 0.36 \times 200 H=72JH = 72 J

Explanation:

The chemical energy in the cell is converted to electrical energy, and as current flows through the resistance, it generates heat according to Joule's law of heating.

Problem 5:

In a laboratory experiment with a Voltaic cell, a voltmeter measures a potential difference of V=1.1VV = 1.1 V. If a resistor of R=5.5ΩR = 5.5 \Omega is connected across the terminals, determine the current (II) flowing through the circuit.

Circuit diagram representation of a voltaic cell with a 5.5 Ohm resistor.

Solution:

Given: Potential Difference (VV) = 1.1V1.1 V Resistance (RR) = 5.5Ω5.5 \Omega

According to Ohm's Law: V=I×RV = I \times R I=VRI = \frac{V}{R} I=1.15.5I = \frac{1.1}{5.5} I=15=0.2AI = \frac{1}{5} = 0.2 A

Explanation:

The potential difference maintained by the chemical reaction in the cell drives the flow of charge through the external resistance.