Two resistors of resistance 1.0 Ω and 2.0 Ω connected in parallel are linked in series to a 3.0 Ω resistor. All three of this is in parallel with a fourth resistor. If the total effective resistance is 1.0 Ω, what is the resistance of the fourth resistor? a)11/3
b)11/8
c)4
d)3

11/8

Explanation:

Given the following:

R1 = 1.0 Ω, R2 = 2.0 Ω, R3 = 3.0 Ω

Total effective resistance = 1.0 Ω

Resistance in parallel :

R = (R1×R2) / R1+R2

Resistance in series :

R = R1 + R2 + +Rn

Therefore,

R1 and R2 in parallel

Re = (1 × 2) / (1 + 2) = 2/3 Ω

Re is linked in series to R3

Rf = Re + R3

Rf = 2/3 + 3 = 11/3

Rf linked in parallel to R4 with effective resistance being 1 Ω

11/ 3 and R4 in parallel gives 1 Ω

Re = (11/3 × R4) / (11/3 + R4)

1 = 11/3 R4 / 11/3 + R4

Cross multiply

11/3 + R4 = 11/3 × R4

11/3 = 11/3R4 - R4

11/3 = 8/3R4

Multiply both sides by 3/8

33/24 = R4

IN LOWEST TERM

R4 = 11/8

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

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