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Chapter 6

Q. 6.15

For the parallel network in Fig. 6.29 (all standard values):
a. Determine the total resistance R_T.
b. Find the source current and the current through each resistor.
c. Calculate the power delivered by the source.
d. Determine the power absorbed by each parallel resistor.
e. Verify Eq. (6.10).

P_E=P_{R_1}+P_{R_2}+P_{R_3}                           (6.10)


Verified Solution

a. Without making a single calculation, it should now be apparent from previous examples that the total resistance is less than 1.6 kΩ and very close to this value because of the magnitude of the other resistance levels:

R_T=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} }=\frac{1}{\frac{1}{1.6k\Omega }+\frac{1}{20k\Omega }+\frac{1}{56k\Omega } }\\[0.5cm] =\frac{1}{625\times 10^{-6}+50\times 10^{-6}+17.867\times 10^{-6}}=\frac{1}{692.867\times 10^{-6}}\\ and   R_T=1.44k\Omega

b. Applying Ohm’s law gives

I_s=\frac{E}{R_T }=\frac{28V}{1.44k\Omega }=19.44mA

Recalling that current always seeks the path of least resistance immediately tells us that the current through the 1.6 kΩ resistor will be the largest and the current through the 56 kΩ resistor the smallest.

Applying Ohm’s law again gives

I_1=\frac{V_1}{R_1}=\frac{E}{R_1}=\frac{28V}{1.6k\Omega }=17.5mA\\[0.5cm] I_2=\frac{V_2}{R_2}=\frac{E}{R_2}=\frac{28V}{20k\Omega }=1.4mA\\[0.5cm] I_3=\frac{V_3}{R_3}=\frac{E}{R_3}=\frac{28V}{56k\Omega }=0.5mA

c. Applying Eq. (6.11) gives

P_E=EI_s      (watts, W)                      (6.11)


d. Applying each form of the power equation gives

P_1=V_1I_1=EI_1=(28V)(17.5mA)=490mW\\[0.5cm] P_2=I^{2}_{2}R_2=(1.4mA)^2(20k\Omega )=39.2mW\\[0.5cm] P_3=\frac{V^{2}_{3} }{R_3}=\frac{E^2}{R_3}=\frac{(28V)^2}{56k\Omega }=14mW

A review of the results clearly substantiates the fact that the larger the resistor, the less is the power absorbed.

e.                          P_E=P_{R_1}+P_{R_2}+P_{R_3}\\[0.5cm] 543.2mW=490mW+39.2mW+14mW=543.2mW           (checks)