Two resistances, R₁ and R₂, are connected in series across a 12-V battery. The current increases by 0.18 A when R₂ is removed, leaving R₁ connected across the battery. However, the current increases by just 0.055 A when R₁ is removed, leaving R₂ connected across the battery. Find (a) R₁ and (b) R₂. R₁ R₁ ww R₂ www R2₂ ww

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Two resistances, R₁ and R₂, are connected in series across a 12-V battery. The current increases by 0.18 A when R₂ is removed, leaving
R₁ connected across the battery. However, the current increases by just 0.055 A when R₁ is removed, leaving R₂ connected across the
battery. Find (a) R₁ and (b) R₂.
(a) Number
(b) Number
R₁
ww
11
remove R₂
Units
Units
R₁
V
◄►
R₂
remove R₁
R₂
www
12
V
Transcribed Image Text:Two resistances, R₁ and R₂, are connected in series across a 12-V battery. The current increases by 0.18 A when R₂ is removed, leaving R₁ connected across the battery. However, the current increases by just 0.055 A when R₁ is removed, leaving R₂ connected across the battery. Find (a) R₁ and (b) R₂. (a) Number (b) Number R₁ ww 11 remove R₂ Units Units R₁ V ◄► R₂ remove R₁ R₂ www 12 V
Expert Solution
Step 1

Given,

Two resistance , R1 and R2 are connected in series across a 12-V battery.

The current increases by 0.18 A when R2 is removed leaving R1 connected across the battery.

The current increases by 0.055 A when R1 is removed leaving R2 connected.

Equivalent of R1 and R2 connected in series is equals to R1+R2.

We know that current = voltageResistance

Let 12 VR1+R2 = i

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