Answers: (a)1.44 Ohm (b) 2.88 W (c) 0.24 A and 0.12 A (d) 1.44 W (top) and 0.35

Question

Answers: (a)1.44 Ohm (b) 2.88 W (c) 0.24 A and 0.12 A (d) 1.44 W (top) and 0.35 W in each of the lower four (f) 6,25 Ohm

8. Resistor Network A network contains five identical resistors connected as shown. Each individual resistor has value R = 25 . (a) Find the equivalent resistance of the network. (b) Suppose the network is connected to a 12 V power supply. Find the total power dissipation. (c) Find the current in each resistor (d) Find the power dissipation in each resistor (e) Find the voltage across each resistor (1) In the circuit shown in the figure, the power dissipation in the upper resistor is different than that in the lower resistors. Suppose we want all resistors to dissipate the same power. If the lower resistors are still 25 , find the value of the upper resistor. (Explain.)

Explanation / Answer

a) Resistorisn in series = 25 + 25 = 50 ohm

50 ohm and 50 ohm are in parallel

R = 50*50 / (50 + 50 ) =25 ohm

both 25 ohm are in series

R(equi) = 25 + 25 = 50 ohm

b) P = V^2 / R = 12^2 / 50 = 2.88 W

c) I = V/R = 12/ 50 = 0.24 A

d) current through outer resitor 0.24 A

current through other resitors = 0.24*50 / (50 + 50 ) =0.12 A

Power through the top = I^2*R = 0.24^2*25 = 1.44 W

Power thorugh others = 0.12^2*25 = 0.36 W

e) voltage across top resistor = IR = 0.24*25 = 6 V

Voltage across other resistors = 0.24*25 = 6 V

f) Power through Upper resistor = I^2 R

0.36 = 0.24^2*R

R = 6.25 ohm

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