Home 2nd Year Physics Notes Class 12 Physics Chapter 13 Short Questions and Numerical Notes – Current Electricity

# Class 12 Physics Chapter 13 Short Questions and Numerical Notes – Current Electricity

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In this post, I am sharing Class 12 Physics Chapter 13 Short Questions and Numerical Notes PDF for the students of Intermediate Part 2. This chapter’s name is Current Electricity. Students can download the Class 12 Physics Chapter 13 Short Questions and Numerical Notes PDF on their laptop or mobile. These Physics Notes are for all the boards working under the Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. Here are the complete FSC 2nd Year Physics Notes PDF chapters.

## FSC Part 2 Physics Chapter 13 Short Questions and Numerical Notes – Current Electricity

Class 12 Physics Notes

#### Why does the resistance of a conductor rise with temperature?

As we know that resistance offerd by a conductor to the flow of current is due to the collisions, of free elecrons with atoms of lattice. As temperature of the conductor rises, the amplititude of vibration of the atoms in the lattice increases and hence the probability of their collisions with free electrons also increases. Hence resistance of conductor rise with temperature.

#### What are the difficulties in testing whether the filament of a lightes bulb obeys Ohm’s law?

According to Ohm’s law current is directly proportional to applied potential difference providing physical state of conductor must remain constant therefore when current passes through the filament of bulb, initially the temperature of filament Oby’s Ohm’s law but with the passage of time, its temperature increases, so resistance of filament increases therefore Ohm’s law is not valid due to increases in temperature.

#### Explain why the terminal potential difference of a battery decreases when the current drawn from it is increased?

We know that the relation between terminal potential difference and emf is

Vt=E-Ir

Here is the internal resistance of cell.

It is clear that when current I is large, the factor Ir becomes large and Vt becomes small. Thus the potential difference of a battery decreases when the current drawn from it increases.

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