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.
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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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
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
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
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