IIT JEE , AIEEE Forum - Physics , Electricity

niki61

a battery of e.m.f. 15v and internal resistance 2 ohm is connected to two resistors of 4 ohm and 6 ohm joined(a)in series (b)in parallel. find in each case the electrical energy spent per minute in 6 ohm resistor

nishant_88

a)in series
total resistance=12 ohm
current=15/12
power=(15/12)²*6
energy spent per minute=power*60 J=562.5 J

b)in parallel
total resistance=6*4/(4+6)+2=4.4 ohm
current=15/4.4
terminal pd=15-current*internal resistance=15-15*2/4.4=36/4.4 V
pd across 6 ohm=terminal pd=36/4.4
power=(pd)²/6
energy spent per minute=power*60 J=669.42 J

edison

Q) a battery of e.m.f. 15v and internal resistance 2 ohm is connected to two resistors of 4 ohm and 6 ohm joined(a)in series (b)in parallel. find in each case the electrical energy spent per minute in 6 ohm resistor.

Sol) (a) In series

R = 2+4+6 = 12 ohms

so I = V/R = 15/12 = 5/4 = 1.25 A

so power consumed by 6 ohm resistor = I² r = 1.25 ² x 6 = 9.375 Watt

so Energy spent in 1 min or 60s = 9.375 x 60 = 562.5 J

(b) for parallel

in 4 and 6 ohm are in parallel then R = 24/10 = 2.4

which is in series with internal resistance of 2 ohm

so R_eff = 4.4 ohm

I = 15/4.4

so energy spent in 60s = 60 x I²R_eff

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