Question

Two resistances, R₁ and R₂, are connected in series across a 12-V battery. The current increases by 0.25 A when R₂ is removed, leaving R₁ connected across the battery. However, the current increases by just 0.10 A when R₁ is removed, leaving R₂ connected across the battery. Find (a) R₁ and (b) R₂.

Answer 1

R1 + R2 = 12 / I

R1 + R2 = 12 / ( I + 0.25 ) + 12 / ( I + 0.1 )

Two equations straight from what you got.

12/ I = 12 / ( I + 0.25 ) + 12 / ( I + 0.1 )

12's cancel

( I + 0.25 ) ( I + 0.1 ) = I (( I + 0.25 ) + ( I + 0.1 ))

Expand out and solve quadratic and get

I = 0.158 A

so R1 = 12/(I+0.25) = 29.41 ohms

R2 = 12/(I+0.1) = 46.51 ohms

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Answer 2

12/r1 = 12/(r1+r2) +0.25

12/r2 = 12/(r1+r2) + 0.10

=> 12(1/r1 -1/r2) = 0.15

=> 1/r1 -1/r2 = 0.0125

=> r1 = 8.48 , r2 = 9.48 ohm

Answer 3

Let current in the circuit be I when both R1 and R2 are connected in series.

From Ohm's law,

Voltage in the circuit(V) = I * (R1 + R2) [As R1 and R2 are connected in series]

=> 12 = I * (R1 + R2) ......(1)

Now as the current increases by 0.25 A when R2 is removed, we have:-

12 = (I+0.25) * R1 ...... (2)

Now as the current increases by 0.1 A when R1 is removed, we have:-

12 = (I+0.1) * R2 ...... (3)

Solving equations 1,2, 3 we get:-

I = 0.158 A

R1 = 29.4035 Ohms

R2 = 46.5116 Ohms

Answer 4

12 = I (R1 + R2)......eq. 1

12 = (I +0.25) * R1.eq.. 2

12 = (I +0.10) * R2..eq..3

by solving we get ....R1 =29.41

R2 = 48.02