Can anyone explain me NCERT class 10 Ch-Electricity example 12.5?

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Yhh i completed this chapter interesting toooo. .. . . type ur question ..

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Yes , I can explain your answer

In the example 12.5 , the resistance of a wire is 26Ω , the diameter is 0.3mm and the length of the wire is 1m.

1m = 1000mm. R = 26Ω
? = 0.3mm. d = 0.3mm

L = 1m
d = 3 × 10-⁴
r = d/2
r = 3 × 10-⁴ / 2
R = p l/a
R×A/l = p. A = R²
( Because wire is in the cylinder form with circular ends)
R = πr²/l = p
26 × 22 × 9 × 10-8 / 7 × 4 = p

Substitution of values in this gives
p = 1.84 × 10-6 Ω m

It is Manganese.

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Type ur question here

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Question type Karo...Book to mene junior ko diya he vacation ke liye

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Bro abhi bas 2 hi din padhai hui hai matlab ek page bhi over nahi hua aur tum Example 12.5 pooch rahe ho. so, sorry bro please forgive me

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This chapter has not been started yet in my scl

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Example 12.5:

Question: A hot plate of an electric iron connected to a 220 V line has two resistance wires A and B in parallel. When a current of 5 A is passed through it, the potential difference across wire A is 200 V and across wire B is 20 V. What is the resistance of wire A and wire B?

Solution:

Given:
- Potential difference across wire A = 200 V
- Potential difference across wire B = 20 V
- Total potential difference (V) = 220 V
- Total current (I) = 5 A

1. Calculating Resistance of Wire A (RA):
- The potential difference (V) across a resistor is directly proportional to the current (I) passing through it, according to Ohm's law (V = IR).
- Therefore, we can write the equation for wire A as: 200 V = RA × 5 A
- Rearranging the equation, we have: RA = 200 V / 5 A
- Solving this equation, we find that the resistance of wire A is 40 Ω.

2. Calculating Resistance of Wire B (RB):
- Similarly, we can write the equation for wire B as: 20 V = RB × 5 A
- Rearranging the equation, we have: RB = 20 V / 5 A
- Solving this equation, we find that the resistance of wire B is 4 Ω.

Conclusion:
- The resistance of wire A is 40 Ω and the resistance of wire B is 4 Ω.

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