Question

Two resistors are connected in series gives an equivalent resistance of 10 Ω. When connected in parallel, gives 2.4 Ω. Then the individual resistance are
(a) each of 5 Ω
(b) 6 Ω and 4 Ω
(c) 7 Ω and 4 Ω
(d) 8 Ω and 2 Ω

Answer 1

Answer:

Explanation:

LET THE TWO EQUIVALENT RESISTANCE BE R1 AND R2

WHEN CONNECTED IN SERIES,

R NET=R1+R2

         =R1+R2=10.................(1)

WHEN CONNECTED IN PARALLEL,

R NET=R1R2/(R1+R2)=2.4 ............(2)

SOLVING 1 AND 2

R1=6 AND R2=4

SHORTCUT

CONDITION (1)    R1+R2=10     FROM THIS (C) CAN BE ELIMINATED AS 7+4≠10

CONDITION (2)

R1R2/(R1+R2)=2.4

R1R2/10=2.4

R1R2=24  

ONLY OPTION C SATISFIES THIS

HOPE IT HELPS.