Question

Two resistances, r1 and r2, are connected in series across a 12-v battery. the current increases by 0.20 a when r2 is removed, leaving r1 connected across the battery. however, the current increases by just 0.13 a when r1 is removed, leaving r2 connected across the battery. find (a) r1 and (b) r2.

Answer 1

The values of resistance r_1=33.3 ohm and r_2 = 41.7 ohm

The total resistance in series is the sum of the value of the resistances. Applying Ohm's law,

`V=I(r_(1) +r_(2)` .....(1)

When the resistor r_1 is connected, the current increases by 0.2 A

Apply Ohm's law for the resistance r_1.

`V=(I+0.2)r_(1)` ......(2)

When the resistor r_2 is connected, the current increases by 0.13 A.

Apply Ohm's law for the resistance r_2.

`V=(I+0.13)r_(2)` ......(3)

From equations (1) and (2)

`image`

`(r_1+r_2)/(r_1) =(I+0.2)/(I) \\ 1+(0.2)/(I) =1+(r_2)/(r_1) \\ (r_2)/(r_1) =(0.2)/(I) ......(4)`

Similarly from equations (1) and (3), it can be seen that

`(r_1+r_2)/(r_2) =(I+0.13)/(I) \\ (r_1)/(r_2) =(0.13)/(I) ..... (5)`

From equations (4) and (5),

I=0.16 A

Substituting in equations (2) and (3)

`r_1 =33.3 Ohm\\ r_2 =41.7 Ohm`