A battery consisting of 10 cells in series has two cells reverse connected by mistake .E.M.F. of each cell is 1.5V and internal resistance 0.1.The value of external resistance is 4.Find the reduction in current due to the two cells being reverse connected.
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WHEN 10 CELLS ARE PROPERLY CONNECTED
Total EMF = 10 x 1.5 = 15 V
Total Circuit Resistance = R + hr
= 4 + 10 x 0.1 = 5 ohm
Circuit Current = 15/5 ➡ 3A
WHEN 2 CELLS ARE REVERSE CONNECTED
Total EMF = ( 8 x 1.5 ) ( 2 x 1.5 ) ➡ 9V
Total circuit resistance = 5 ohm ( same as before )
Circuit current = 9/5 ➡ 1.8A
HENCE , THE REDUCTION IN CURRENT
➡ 3 - 1.8 = 1.2 A ( Answer )
Total EMF = 10 x 1.5 = 15 V
Total Circuit Resistance = R + hr
= 4 + 10 x 0.1 = 5 ohm
Circuit Current = 15/5 ➡ 3A
WHEN 2 CELLS ARE REVERSE CONNECTED
Total EMF = ( 8 x 1.5 ) ( 2 x 1.5 ) ➡ 9V
Total circuit resistance = 5 ohm ( same as before )
Circuit current = 9/5 ➡ 1.8A
HENCE , THE REDUCTION IN CURRENT
➡ 3 - 1.8 = 1.2 A ( Answer )
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