krit.club logo

Electricity - ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE11.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE11.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE11.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE11.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE

Grade 10CBSE

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

🔑Concepts

•

Electric potential difference (VV) between two points in an electric circuit is defined as the work done (WW) to move a unit charge (QQ) from one point to the other.

•

The SI unit of electric potential difference is the volt (VV), named after Alessandro Volta (17451745–18271827).

•

One volt is the potential difference between two points in a current-carrying conductor when 11 joule of work is done to move a charge of 11 coulomb from one point to the other.

•

Potential difference is measured by an instrument called the voltmeter, which is always connected in parallel across the points between which the potential difference is to be measured.

•

For electrons to flow in a conductor, there must be a difference in electric pressure, called the potential difference, which is usually maintained by a cell or a battery.

📐Formulae

V=WQV = \frac{W}{Q}

1 V=1 J1 C1\text{ V} = \frac{1\text{ J}}{1\text{ C}}

W=V×QW = V \times Q

💡Examples

Problem 1:

How much work is done in moving a charge of 2 C2\text{ C} across two points having a potential difference 12 V12\text{ V}?

Solution:

W=24 JW = 24\text{ J}

Explanation:

Given: Charge Q=2 CQ = 2\text{ C} and Potential Difference V=12 VV = 12\text{ V}. Using the formula for work done, W=V×QW = V \times Q, we substitute the values: W=12 V×2 C=24 JW = 12\text{ V} \times 2\text{ C} = 24\text{ J}.

Problem 2:

If 60 J60\text{ J} of work is required to move a certain amount of charge between two points where the potential difference is 6 V6\text{ V}, calculate the magnitude of the charge.

Solution:

Q=10 CQ = 10\text{ C}

Explanation:

Given: Work done W=60 JW = 60\text{ J} and Potential Difference V=6 VV = 6\text{ V}. Since V=WQV = \frac{W}{Q}, we can rearrange the formula to find charge: Q=WVQ = \frac{W}{V}. Substituting the values, Q=60 J6 V=10 CQ = \frac{60\text{ J}}{6\text{ V}} = 10\text{ C}.