Two resistances, R 1 and R 2 , are connected in series across a 12-V battery. Th

Question

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.17  A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.11  A when R1 is removed, leaving R2 connected across the battery. Find (a) R1 and (b) R2.

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.17 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.11 A when R1 is removed, leaving R2 connected across the battery. Find (a) R1 and (b) R2.

Explanation / Answer

R1 + R2 = 12 / I

R1 + R2 = 12 / ( I + 0.17 ) + 12 / ( I + 0.11 )

12/ I = 12 / ( I + 0.17 ) + 12 / ( I + 0.11 )

( I + 0.17 ) ( I + 0.11 ) = I (( I + 0.17 ) + ( I + 0.11 ))

0.28I+ I^2+ 0.0187=2I^2 +0.28I

I=0.13675

R1=39.12

R2=48.63

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