EMF of an electric cell is 2.1 V and its internal resistance is 0.1 ohm. When the two poles of the cell are connected with an external resistance, a potential difference of 2.08V is obtained. determine the value of the external resistance.<br /><br />PLZ HELP ME FRNDZZZ. ...NO SPAM.....PLZZZZ.....
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Howdy!!
your answer is ---
the resistance of cell r =0.1 ohm
& emf = 2.1V
let, external resistance is R
we know that emf is equal to the potential difference when no current is flow
also, E = IR + Ir
=> 2.1 = V + I × 0.1
[ since potential difference = IR ]
=> 2.1 = 2.08 + I × 0.01
=> I = 0.02/0.1 = 0.2 ampere
now ,
V = IR
=> 2.08 = 0.2 × R
=> R = 2.08/0.2 = 10.4ohm
hence , external resistance is 10.4 ohm
hope it help you
your answer is ---
the resistance of cell r =0.1 ohm
& emf = 2.1V
let, external resistance is R
we know that emf is equal to the potential difference when no current is flow
also, E = IR + Ir
=> 2.1 = V + I × 0.1
[ since potential difference = IR ]
=> 2.1 = 2.08 + I × 0.01
=> I = 0.02/0.1 = 0.2 ampere
now ,
V = IR
=> 2.08 = 0.2 × R
=> R = 2.08/0.2 = 10.4ohm
hence , external resistance is 10.4 ohm
hope it help you
Rupsa1111:
but the given answer us 10.4 ohm
sryy
now see my answers
thank u soooo much
welcome
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