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Chapter 6

Q. 6.13

For the parallel network in Fig. 6.23.
a. Find the total resistance.
b. Calculate the source current.
c. Determine the current through each branch.

Step-by-Step

Verified Solution

a. Applying Eq. (6.3) gives

R_T=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdot \cdot \cdot +\frac{1}{R_N} }                  (6.3)

 

R_T=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} }=\frac{1}{\frac{1}{10\Omega }+\frac{1}{220\Omega }+\frac{1}{1.2k\Omega } }\\[0.5cm] =\frac{1}{100\times 10^{-3}+4.545\times 10^{-3}+0.833\times 10^{-3}}=\frac{1}{105.38\times 10^{-3}}\\[0.5cm] R_T=9.49\Omega

Note that the total resistance is less than that of the smallest parallel resistor, and its magnitude is very close to the resistance of the smallest resistor because the other resistors are larger by a factor greater than 10:1.

b. Using Ohm’s law gives

I_s=\frac{E}{R_T}=\frac{24V}{9.49\Omega }=2.53A

c. Applying Ohm’s law gives

I_1=\frac{V_1}{R_1}=\frac{E}{R_1}=\frac{24V}{10\Omega }=2.4A\\[0.5cm] I_2=\frac{V_2}{R_2}=\frac{E}{R_2}=\frac{24V}{220\Omega }=0.11A\\[0.5cm] I_3=\frac{V_3}{R_3}=\frac{E}{R_3}=\frac{24V}{1.2K\Omega }=0.02A