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Electricity: Magnetic and Heating Effects - Rechargeable batteries

Grade 8CBSE

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

🔑Concepts

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Heating Effect of Electric Current: When an electric current flows through a high-resistance wire (like nichrome), the wire becomes hot. This is known as the heating effect. The heat produced HH depends on the current II, resistance RR, and time tt.

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Electric Fuse: A safety device based on the heating effect of current. It consists of a wire with a low melting point that melts and breaks the circuit if the current exceeds a safe limit.

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Magnetic Effect of Electric Current: When electric current flows through a wire, it behaves like a magnet. This was discovered by Hans Christian Oersted. This effect is used to create electromagnets.

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Electromagnet: A coil of insulated wire wound around a soft iron core. It acts as a magnet only when current flows through it. Its strength can be increased by increasing the number of turns nn or the current II.

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Primary vs. Secondary Cells: Primary cells (like dry cells) cannot be recharged as their chemical reaction is irreversible. Secondary cells (rechargeable batteries) can be recharged by passing current through them in the opposite direction, reversing the chemical reactions.

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Rechargeable Batteries: Examples include Lead-acid batteries (used in cars) and Lithium-ion batteries (used in mobile phones). They convert electrical energy into chemical energy during charging and vice versa during discharging.

📐Formulae

H=I2RtH = I^2 R t

V=I×RV = I \times R

P=V×IP = V \times I

Total Resistance (Series)=R1+R2+R3...\text{Total Resistance (Series)} = R_1 + R_2 + R_3...

💡Examples

Problem 1:

Calculate the total resistance in a circuit where two heating elements with resistances of 55Ω55 \Omega and 38Ω38 \Omega are connected in series.

Solution:

Rtotal=R1+R2R_{total} = R_1 + R_2 55+3893\begin{array}{r} 55 \\ +38 \\ \hline 93 \end{array} So, total resistance is 93Ω93 \Omega.

Explanation:

In a series circuit, the individual resistances are added together to find the total resistance.

Problem 2:

An electric heater is rated at 1500W1500 W and operates at 250V250 V. Find the current II drawn by the heater.

Solution:

Using the formula P=V×IP = V \times I, we can derive I=PVI = \frac{P}{V}. I=1500250I = \frac{1500}{250} I=6AI = 6 A

Explanation:

The current is calculated by dividing the power (in Watts) by the voltage (in Volts).

Problem 3:

A rechargeable battery is being charged. If it receives a constant current of 2A2 A for 55 hours, calculate the total charge QQ stored (given Q=I×tQ = I \times t, and 1 hour=3600 seconds1 \text{ hour} = 3600 \text{ seconds}).

Solution:

t=5×3600=18000 st = 5 \times 3600 = 18000 \text{ s} Q=I×tQ = I \times t Q=2×18000=36000 CQ = 2 \times 18000 = 36000 \text{ C}

Explanation:

The total charge QQ is the product of current and time in seconds. 11 Coulomb (1C1 C) is the charge transferred by 1A1 A in 1s1 s.

Rechargeable batteries Class 8 Notes & Examples