Question

Two terminals of a cell are marked as x (+6v) and y (-6v). the cell is connected to a wire of resistance r (2 ohm). calculate work done in carrying a charge of 2c from:-
(i) x to y along the wire r.
(ii) y to x through the cell.

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Answer 1

ANSWER

Applying Kirchoff's Voltage Law in the loop,

2I−6+12+3I=0

⟹5I=−6

⟹I=−1.2A

P.D across 6V battery =6−2I=6+2.4=8.4V

hope it helps

Answer 2

i) x to y -

work done = v x q

so workdone in moving 2c from x to y = 6 x 2 = 12

ii) y to x -

its inside the cell , so the current is flowing from -6 to +6

potential difference = -6-(+6) = -12

workdone = v x q = -12 x 2 = -24

this maybe correct , although i aint sure!!

if there's any correction please comment!! i will rectify!!

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