## 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 $$ \frac{1}{80 \Omega}=n\left(\frac{1}{40 \Omega}\right) \quad \text { or } \quad n=5 $$

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