Question

the given circuit, A, B, C and D are four lamps connected with a battry of 60v
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Answer 1

so, R=V/I lamp1 , R= 60/3 = 20 lamp2 , R= 60/4 = 15 lamp3 , R= 60/5 = 12 lamp4 , R= 60/3 = 20 therefore lamp 3 which has less resistance will glow brightest . 4: we know that total resistance in parallel circuit is 1÷total resistance = 1/R1 +1/R2+1/R3+

thus 1/total resistance = 1/60/3 + 1/60/4 + 1/60/5 + 1/60/3 1/R total = 15/60 R total = 60/15 R total = 4 in-the-given-circuit-a-b-c-and-d-are-four-lamps-connected-with-a-battery-of-60v

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