If there are two resistors r1 and r2 in a parallel circuit. r2 is 15 ohms greater than r1 and R=4 ohms calculate r1 and r2.

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

If 2 resisters  R1 and R2 ohms are connected in parallel, then the resulting resistance R of the circuit is given by the relation:

1/R = 1/R1 +1/R2 ....(1)

Inthe given situation, R2 = 15+R1 ohms. The  circuit resistance R = 4ohms.

We substitute the values in the relation (1) and find the resixstance R1:

1/4 = 1/R1 + 1/(R1+15).

Multiply both sides by 4R1(R1+1.5) to get rid of the denominators:

R1(R1+15) = 4(R1+15) +4R1

R1^2+15R1 = 4R1+60+4R1 = 8R1+60

R1^2+15R1-8R1- 60 = 0

R1^2 +7R1 - 60 = 0 is a quadratic equation in R1

We solve by the formula of quadratic equation:

R^2+12R1-5R1-60 = 0

R1(R1+12) -5(R1+12) = 0

(R1+12)(R1-5) = 0

Therefore R1 = 5 ohm  and  R2 = 5+15 = 20 oHm.

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

For resistances in parallel the equivalent resistance of two resistors r1 and r2 is given as the reciprocal of the sum of their inverses.

Or if the equivalent resistance is R. 1/R = 1/r1 + 1/r2.

Now it is given that r2 = 15 greater than r1 or 15 + r1.

Also the equivalent resistance R = 4

=> 1/4 = 1/ r1 + 1/(15 + r1)

=> 1/4 = (15 + r1 + r1) / [r1 *(15+r1)]

=> [r1 *(15+r1)] = 4*(2*r1 + 15)

=> [r1 *(15+r1)] = 8 r1 + 60

=> 15r1 + r1^2 = 8r1 + 60

=> r1^2 + 7r1 - 60 =0

=> r1^2 + 12r1 - 5 r1 - 60=0

=> r1( r1+12) - 5(r1- 12)=0

=> (r1 -5)(r1+12) =0

r1 can be -12 or 5

Now a resistance cannot be negative. Therefore r1= 5.

r2 = 5+ 15 = 20

So the resistance r1 is 5 ohms and r2 is 20 ohms.

The required result is 5 ohms and 20 ohms

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

R is the joint resistance and it is calculated for a parallel circuit as:

1/R = 1/r1 + 1/r2

We know, from enunciation, that r2 = r1 + 15

We'll substitute the known value into the formula of joint resistance:

1/4 = 1/r1 + 1/(r1+15)

We'll calculate the LCD = 4r1(r1+15)

r1(r1+15) = 4(r1+15) + 4r1

We'll remove the brackets:

r1^2 + 15r1 = 4r1 + 60 + 4r1

We'll combine like terms:

r1^2 + 15r1 = 8r1 + 60

We'll subtract 8r1 + 60:

r1^2 + 15r1 - 8r1 - 60 = 0

r1^2 + 7r1 - 60 = 0

We'll apply the quadratic formula:

r1 = [-7+sqrt(49+240)]/2

r1 = (-7+17)/2

r1 = 5

or

r1 = (-7-17)/2

r1 = -24/2

r1 = -12 is not acceptable.

The value of r2 = r1 + 15

r2 = 5 + 15

r2 = 20

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