Please answer #2 (#1 is only a reference for #2). 2. Replace all the capacitors

Question

Please answer #2 (#1 is only a reference for #2).

2. Replace all the capacitors of the previous paper HW #06 by resistors, keep the battery voltage the same, and let the resistors have resistance R,-1 k. Find everything: Vi, li, Passi,1-6. Organize your results in a nice table. Assume all the resistors are ohmic bulbs. How would you rearrange the bulbs to make the circuit as bright as possible? How would you rearrange the bulbs to make the circuit as dim as possible? (All bulbs must glow; also the circuit topology may not be altered!) Assume brightness is proportional to the power dissipated.

Explanation / Answer

Resistors R3 and R4 are in series .So R3+R4 = 7kohms

These are in parallel with R5 so -total Eq resistance = 7kohms X 5 Kohms/12 Kohms = (35/12) kohms

The above (35/12) kohm is in series with 6kOhm, so total resistance = [(35/12) +6] kohm = (107/12)Kohmm

Finally, (107/12) Kohm is parallel with 2 ohms, so total resistance = (107/12) X 2 /(107/12+2) = (214/12)/(131/12)

= (214/131) kohms  

This resistor is in series with R1 , R (Eq) = (214/131) +1 = (345/131)Kohms

So current leaving battery = 24V/ (345/131)Kohms = 9.11 X 10^(-3) A = 9.11 milli A = i1

I1 = 9.11mA ; V1 = 9.11 mA X 1Kohm = 9.11 V , P1 = 9.11V X 9.11 mA = 83.0 milli watts

i2 = 14.88/2k ohm = 7.44 mA ; v2 = 14.88V ; P2 = 14.88v X 7.44 mA = 110.7mW

i1-i2 = 9.11-7.44 = 1.67 mA

V6 = 1.67 mA X 6 Kohm = 10.02 V ; i6= 1.67mA ; P6 = 167mW

V5 = 14.88-10.02 = 4.86V ; i5= 4.86/5Kohm = 0.97 mA; P5 = 4.72mW

i3 = 1.67 -0.97 = 0.7 mA; V3= 2.1 V ; P3 = 1.47mW

i4 = 0.7 mA ; V3= 2.8V ; P3 = 1.96 mW

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