Question

Two resistances, R1 and R2, are connected in series across a 9-V battery. The current increases by 0.450 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.225 A when R1 is removed, leaving R2 connected across the battery. a) Find R1 b) Find R2

Answer 1

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

Step-by-step explanation:

Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.

When the current increases by 0.450 A wen only R1 is in the circuit, the current is

9/R1 + R2 + 0.450 A = 9/R1 (1)

When the current increases by 0.225 A when only R2 is in the circuit, the current is

9/R1 + R2 + 0.225 A = 9/R2 (2)

equation (1) - (2) equals

9(1/R1 - 1/R2) = 0.450 A - 0.225

9(1/R1 - 1/R2) = 0.125

(1/R1 - 1/R2) = 0.125 A/9 = 0.0138

1/R1 = 0.0138 + 1/R2

R1 = R2/(1 + 0.0138R2) (3)

From (1)

9/R1 - 9/R1 + R2 = 0.450 A

9R2/[R1(R1 + R2)] = 0.450 A

R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5

R2/[R1(R1 + R2)] = 0.5 (4)

From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,

(1 + 0.0138R2) = 0.5(R1 + R2)

0.5R1 + 0.5R2 = 1 + 0.0138R2

0.5R1 = 1 + 0.0138R2 - 0.5R2

0.5R1 = 1 - 0.4862R2 (5)

Substituting (3) into (5) we have

0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2

R2 = (1 + 0.0138R2)(1 - 0.4862R2)

R2 = 1 - 0.4724R2 - 0.0067R2²

Collecting like terms, we have

0.0067R2² + 0.4724R2 + R2 - 1 = 0

0.0067R2² + 1.4724R2 - 1 = 0

Using the quadratic formula,

`R_(2) = \frac{-1.4724 +/-\sqrt{(1.4724)^(2) - 4 X 0.0067 X -1} }{2 X 0.0067} \\= (-1.4724 +/-√(2.1680 + 0.0268) )/(0.0268)\\= (-1.4724 +/-√(2.1948) )/(0.0268)\\= (-1.4724 +/- 1.4815 )/(0.0268)\\= (-1.4724 + 1.4815 )/(0.0268) or (-1.4724 - 1.4815 )/(0.0268)\\= (0.0091 )/(0.0268) or (-2.9539)/(0.0268)\\= 0.340 or -110.22`

We choose the positive answer.

So R2 = 0.340 Ω

From (5)

R1 = 0.5 - 0.9931R2

= 0.5 - 0.9931 × 0.340

= 0.5 - 0.338

= 0.162 Ω

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω