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nika2105 [10]
2 years ago
8

\mathrm{question \hookleftarrow}}" alt=" \huge{ \mathrm{question \hookleftarrow}}" align="absmiddle" class="latex-formula">
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
​
Physics
2 answers:
gulaghasi [49]2 years ago
7 0

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

lakkis [162]2 years ago
5 0

Answer:

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

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

<em>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:</em>

<em>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 : = </em><em><u>11/300 R1 = 27.27 ohms</u></em>

Explanation:

<em>I </em><em>think</em><em> </em><em>this</em><em> is</em><em> your</em><em> answer</em><em> </em><em>hope</em><em> it</em><em> will</em><em> help</em><em> you</em><em> have</em><em> a</em><em> great</em><em> day</em><em> bye</em><em> </em><em>:</em><em>)</em>

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Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

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Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

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If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

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f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

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(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

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