Question

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Calculate equivalent resistance of two resistors R1 and R2 in parallel where,

$\sf R_ 1 = (6 \pm0.2 )\: \: ohms$
$\sf R_ 2 = (3 \pm0.1 )\: \: ohms$
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Answer 1

Answer:

Let R is the equivalent resistance for parallel combination and ΔR is the error(in its measurement respective to the error in the values of resistors)

Hence,

R/1 =1/R¹ + 1/R

R/1 = R¹+R²/R¹+ R²

R= 3+6/3×6

∴R=2Ω

Now,

Δ R /R²=ΔR¹/ (R 1 )²+ΔR²+ (R2)²

ΔR /R²=0.1/(3) +0.5(6)²

ΔR = 4

0.1ΔR=0.1

Hence the resistance of the combination is

R=(2.00±0.1)Ω

Answer 2

Answer:

For a parallel circuit with two resistors, the total resistance is calculated from the expression:

For a parallel circuit with two resistors, the total resistance is calculated from the expression:1/R=1/R1+1/R2

For a parallel circuit with two resistors, the total resistance is calculated from the expression:1/R=1/R1+1/R2We are given the total resistance, R, which is 20 ohms and R2 which is 75 ohms. We calculate R1 as follows:

For a parallel circuit with two resistors, the total resistance is calculated from the expression:1/R=1/R1+1/R2We are given the total resistance, R, which is 20 ohms and R2 which is 75 ohms. We calculate R1 as follows:1/20 = 1/R1+1/75 1/R1 : = 11/300 R1 = 27.27 ohms

Explanation:

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