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Electricity: Magnetic and Heating Effects - Dry cells

Grade 8CBSE

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

🔑Concepts

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A dry cell is a primary chemical cell that converts chemical energy into electrical energy. It is called 'dry' because its electrolyte is in the form of a moist paste rather than a liquid, making it portable and leak-proof.

Internal structure of a dry cell showing zinc anode and carbon cathode with paste electrolyte.
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The components of a dry cell include a Zinc container (negative terminal), a Carbon rod (positive terminal), and a mixture of Ammonium Chloride (NH4ClNH_4Cl) paste which acts as the electrolyte.

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When multiple dry cells are connected in series, the positive terminal of one cell is connected to the negative terminal of the next. The total voltage VtotalV_{total} is the sum of individual voltages: Vtotal=V1+V2+⋯+VnV_{total} = V_1 + V_2 + \dots + V_n

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The heating effect in a circuit powered by a dry cell occurs because the electrical energy is converted into thermal energy as current II passes through a resistance RR over time tt, given by H=I2RtH = I^2 R t.

📐Formulae

H=I2RtH = I^2 R t

V=IRV = I R

P=VIP = V I

💡Examples

Problem 1:

A current of 3 A3\text{ A} flows through an electric heater with a resistance of 20 \Omega20\text{ \Omega} for 10 seconds10\text{ seconds}. Calculate the total heat produced in Joules.

Solution:

H=I2RtH = I^2 R t H=(3)2×20×10H = (3)^2 \times 20 \times 10 H=9×20×10H = 9 \times 20 \times 10 H=1800 JH = 1800\text{ J}

Explanation:

By applying Joule's Law of Heating, we square the current, multiply it by the resistance, and then by the time duration to find the energy dissipated as heat.

Problem 2:

Calculate the total voltage required to push a current of 0.5 A0.5\text{ A} through a circuit with a total resistance of 440 \Omega440\text{ \Omega}.

Solution:

V=I×RV = I \times R V=0.5×440V = 0.5 \times 440 V=220 VV = 220\text{ V}

Explanation:

Using Ohm's Law, the potential difference (VV) is the product of the current (II) and the resistance (RR).

Problem 3:

Represent the subtraction of two resistance values 500 \Omega500\text{ \Omega} and 125 \Omega125\text{ \Omega} to find the difference in resistance.

Solution:

500−125375\begin{array}{r} 500 \\ - 125 \\ \hline 375 \end{array}

Explanation:

The difference in resistance between the two components is 375 \Omega375\text{ \Omega}.

Problem 4:

A dry cell with a potential difference of 1.5 V1.5\text{ V} is connected to a bulb with a resistance of 6 \Omega6\text{ \Omega}. Find the current II flowing through the bulb.

Single 1.5V dry cell circuit component.

Solution:

Given: Voltage V=1.5 VV = 1.5\text{ V} Resistance R=6 \OmegaR = 6\text{ \Omega}

Using Ohm's Law: I=VRI = \frac{V}{R} I=1.56I = \frac{1.5}{6} I=0.25 AI = 0.25\text{ A}

The current flowing through the bulb is 0.25 A0.25\text{ A}.

Explanation:

The current is calculated by dividing the electromotive force (voltage) of the dry cell by the total resistance of the circuit.

Problem 5:

Two dry cells of 1.5 V1.5\text{ V} each are connected in series to a resistor. If the total resistance is 2 \Omega2\text{ \Omega} and the current flows for 10 seconds10\text{ seconds}, calculate the heat produced. Use vertical multiplication for the final energy calculation.

Two dry cells connected in series to form a 3V battery.

Solution:

Total Voltage V=1.5 V+1.5 V=3.0 VV = 1.5\text{ V} + 1.5\text{ V} = 3.0\text{ V} Current I=VR=32=1.5 AI = \frac{V}{R} = \frac{3}{2} = 1.5\text{ A}

Heat produced H=I2RtH = I^2 R t H=(1.5)2×2×10H = (1.5)^2 \times 2 \times 10 H=2.25×20H = 2.25 \times 20

2.25×2045.00\begin{array}{r} 2.25 \\ \times 20 \\ \hline 45.00 \end{array}

Total heat produced is 45 J45\text{ J}.

Explanation:

First, the total voltage of the battery is found by summing the cell voltages. Then, the current is determined. Finally, Joule's law of heating is applied to find the energy in Joules.