Answer to Question #266311 in Electric Circuits for Nishat

Question #266311

The equivalent resistance of two resistors R1 and R2 is given by 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 and 𝑅𝑒𝑞 =



𝑅1𝑅2



𝑅1+𝑅2



when they are connected in series and parallel respectively. If 𝑅1 = 22.0 Ω ±



5% 𝑎𝑛𝑑 𝑅2 = 47.0 ± 5% , find the maximum error in the calculation of their equivalent



resistance when they are connected in series and parallel.

1
Expert's answer
2021-11-15T17:50:35-0500

Rs=R1+R2R_s=R_1+R_2

Now errors

∆Rs =(∆R1+∆R2 )

Rs=±(0.05+0.05)=±0.10∆R_s=\pm({0.05}+{0.05})=\pm0.10

Rs=R1+R2=22+48=70ΩR'_s=R_1+R_2=22+48=70\Omega

Now

Req=70±0.10R_{eq}=70\pm0.10

1Req=1R1+1R2\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}

ReqReq=R1R1+R2R2\frac{∆R_{eq}}{R_{eq}}=\frac{∆R_1}{R_1}+\frac{∆R_2}{R_2}

ReqReq=±(0.0522+0.0547)\frac{∆R_{eq}}{R_{eq}}=\pm(\frac{0.05}{22}+\frac{0.05}{47})


ReqReq=±(0.00227+0.00106)=±(0.00303)\frac{∆R_{eq}}{R_{eq}}=\pm(0.00227+0.00106)=\pm(0.00303)

Req=R1R2R1+R2R_{eq}=\frac{R_1R_2}{R_1+R_2}

Req=22×4722+47=14.98ΩR_{eq}=\frac{22\times47}{22+47}=14.98\Omega


Req=14.98×0.00303=0.04988Ω∆R_{eq}=14.98\times0.00303=0.04988\Omega

Req=14.98±0.04988ΩR_{eq}=14.98\pm0.04988\Omega


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