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Q.

Three resistors having resistances of 1.6 Ω, 2.4 Ω, 4.8 Ω are connected in parallel to a 28V battery that has negligible resistance. The current through the battery is

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a

40 A

b

35 A

c

30 A

d

45 A

answer is B.

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Detailed Solution

1Rp=1R1+1R2+1R3

=11.6+12.4+14.8

=1016+1024+1048

=108[12+13+16]

=54[3+2+16]

=54×66=54

1Rp=45

V=IR

I=VRp=2845=28×54⇒I=35A

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