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## Q. 6.8

Determine the total resistance for the parallel combination in Fig. 6.7 using two applications of Eq. (6.5).

$R_T=\frac{R_1R_2}{R_1+R_2}$               (6.5) ## Verified Solution

First the 1 Ω and 4 Ω resistors are combined using Eq. (6.5), resulting in the reduced network in Fig. 6.12:

Eq. (6.4): $R^{\prime }_T=\frac{R_1R_2}{R_1+R_2}=\frac{(1\Omega )(4\Omega )}{1\Omega +4\Omega }=\frac{4}{5}\Omega =0.8\Omega$

Then Eq. (6.5) is applied again using the equivalent value:

$R_T=\frac{R^{\prime }_TR_3}{R^{\prime }_T+R_3}=\frac{(0.8\Omega )(5\Omega )}{0.8\Omega +5\Omega }=\frac{4}{5.8}\Omega =0.69\Omega$

The result matches that obtained in Example 6.3. 