Resistance

Electricity and Magnetism Level 2

The e.m.f of a cell is $1.4 ~ V$ and its internal resistance is $0.2~ Ω.$ What is the terminal potential difference of the cell if the terminals are connected by a wire of resistance $2.6~ Ω?$

1.0 ~ V

1.3 ~ V

2.4 ~ V

1.5 ~ V

2.0 ~ V

1 solution

Rishu Jaar
Nov 1, 2017

Please rotate the image.

Munem Shahriar - 3 years, 7 months ago

Can you help me how, I tried but my phone doesn't allow it or the site , maybe.

Rishu Jaar - 3 years, 7 months ago

@Calvin Lin , sir please help.

Rishu Jaar - 3 years, 7 months ago

Do you have any picture editor on your mobile?

However, here is the rotated image.

Picture link: https://ds055uzetaobb.cloudfront.net/uploads/MstTelzGlO-a52c981f83c5ee20a83857f72aa495bd24bdaa41.jpg

Munem Shahriar - 3 years, 7 months ago

@Munem Shahriar – Thanks now its perfect.

Rishu Jaar - 3 years, 7 months ago

Hey, could you explain the steps from " Now, in case of discharging......"

Aman thegreat - 3 years, 7 months ago

In a 'real' cell/battery(E,r) , the terminal potential difference is V = E - ir , where E is the emf , r is internal resistance and i is the current in the circuit ; when it is discharging(acts as a source or power provider) , which is entirely based on the fact that voltage drops by an amount it when the current exits the battery , but in the case of charging(acts as a load or power consumer) , the terminal p.d is V=E+ir . For more reference Read This , hope it helps . $\ddot\smile$

Rishu Jaar - 3 years, 7 months ago

Resistance

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