For what value of R will the net resistance o | vedclass

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(C) First,simplify the inner part of the circuit. The two $10 \, \Omega$ resistors in parallel give $R_p = \frac{10 	imes 10}{10 + 10} = 5 \, \Omega$

Vedclass · Accepted Answer

$16 \, \Omega$

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(C) First,simplify the inner part of the circuit. The two $10 \, \Omega$ resistors in parallel give $R_p = \frac{10 	imes 10}{10 + 10} = 5 \, \Omega$.This $5 \, \Omega$ is in series with the $10 \, \Omega$ resistor,giving $R_s = 5 + 10 = 15 \, \Omega$.This $15 \, \Omega$ is in parallel with the other $10 \, \Omega$ resistor,giving $R_{p'} = \frac{15 	imes 10}{15 + 10} = \frac{150}{25} = 6 \, \Omega$.Now,the circuit consists of a $10 \, \Omega$ resistor in series with the parallel combination of $R$ and $(10 + 6) \, \Omega = 16 \, \Omega$.Given the total resistance $R_{eq} = 18 \, \Omega$,we have:$18 = 10 + \frac{R 	imes 16}{R + 16}$$8 = \frac{16R}{R + 16}$$8(R + 16) = 16R$$8R + 128 = 16R$$8R = 128$$R = 16 \, \Omega$.

For what value of $R$ will the net resistance of the circuit be $18 \, \Omega$?

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