#3 Consider the figure shown below. There is a battery supplying a voltage V = 1

Question

#3 Consider the figure shown below. There is a battery supplying a voltage V = 130 Volts. The resistance of each resistor is as follows: R_1 = 20 Ohm. R_2 = 100 Ohm and R_3 = R_4 = 66.66 Ohm. As Cite circuit is running a disaster occurs due to a defect in R_4 and the resistor explodes in flames. It is not completely broken and the circuit can be treated as if it was not there. If R_3 can only withstand 120 W of power before failing, will R_3 fail? Explain your reasoning. You are charged by your boss to fix this circuit by replacing the missing R_4 but you only have 200 Ohm resistors at your disposal. Is it possible? What is the combination needed?

Explanation / Answer

here R3 and R4 are connected in parallel

R34 = R3*R4/(R3+ R4)

= 66.66*66.66/(66.66+66.66)

= 33.33 ohm

this combination is connected in series with R1

R134 = R1 + R34

= 20 + 33.33

= 53.33 ohms

this combination is in parallel with R2.

so,

Rnet = R2*R134/(R2 + R134)

= 100*53.33/(100+53.33)

= 34.78 ohms

current through the battery, I = V/Rnet

= 130/34.78

= 3.738 A

current through R1, I1 = I*R2/(R2 + R134)

= 3.738*100/(100 + 53.33)

= 2.438 A

current through R3, I3 = I1/2 (other half of the current passes through R4)

= 1.219 A

ower disspated at R3, P3 = I3^2*R3

= 1.219^2*66.66

= 99 Watts

power dissipted at R3(99) < 120 W

so, it can with stand. <<<<<<----------Answer

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