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Several $40-\Omega$ resistors are to be connected so that 15 A flows from a $120-V$ source. How can this be done? The equivalent resistance must be such that 15 A f...

Question

Several $40-\Omega$ resistors are to be connected so that 15 A flows from a $120-V$ source. How can this be done? The equivalent resistance must be such that 15 A flows from $120 \mathrm{~V}$. Thus, $$ R_{\mathrm{eq}}=\frac{V}{I}=\frac{120 \mathrm{~V}}{15 \mathrm{~A}}=8 \Omega $$ The resistors must be in parallel, since the combined resistance is to be smaller than any of them. If the required number of $40-\Omega$ resistors is $n$, then $$ \frac{1}{80 \Omega}=n\left(\frac{1}{40 \Omega}\right) \

Several $40-\Omega$ resistors are to be connected so that 15 A flows from a $120-V$ source. How can this be done? The equivalent resistance must be such that 15 A flows from $120 \mathrm{~V}$. Thus, $$ R_{\mathrm{eq}}=\frac{V}{I}=\frac{120 \mathrm{~V}}{15 \mathrm{~A}}=8 \Omega $$ The resistors must be in parallel, since the combined resistance is to be smaller than any of them. If the required number of $40-\Omega$ resistors is $n$, then $$ \frac{1}{80 \Omega}=n\left(\frac{1}{40 \Omega}\right) \quad \text { or } \quad n=5 $$



Answers

If $ R $ is the total resistance of three resistors, connected in parallel, with resistances $ R_1 $, $ R_2 $, $ R_3 $, then $$ \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} $$ If the resistances are measured in ohms as $ R_1 = 25 \Omega $, $ R_2 = 40 \Omega $, and $ R_3 = 50 \Omega $, with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of $ R $.


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