Anonymous
Anonymous asked in Science & MathematicsPhysics · 6 months ago

# find voltage across resistor?

determine

the voltage across the current source?

voltage across R4?

voltage across R5? Relevance

first get total resistance as seen by the source.

(in kΩ and mA)

R3+R6 = 6.8+0.39 = 7.19

that in parallel with R5 = 7.19•12(7.19+12) = 4.50

that in series with R2 = 8.2+4.5= 12.7

that in parallel with R4 = 12.7•1/13.7 = 0.927

in series with R1 that is = 3.13

1. E = IR = 3.13 x 790 = 2470 volts

2. voltage across R4 = 790 x 0.927 = 732 volts

3. current thru R2 = 732 / 12.7 = 57.6 mA

voltage across R5 = 57.6 x 4.5 = 259 volts

PS, these are strange values... total power is 1951 watts

PPS, ohms law can be taken as

voltage in volts = current in mA x resistance in kΩ

• Label the node above the 1kΩ node V1 and above the 12Ω node V2. All resistances in kΩ and currents in mA

By KCL the currents leaving V1 = 0:

-790+V1/1+(V1-V2)/8.2 = 0 -----> 6478 = 9.2V1 - V2

V2 = (9.2*V1 - 6478)

By KCL the currents leaving V2 = 0:

(V2-V1)/8.2 + V2/12 + V2/(0.39+6.8)

V2(1/8.2+1/12+1/7.19) = V1/8.2

V2 = (0.354*V1)

Now equate the two equations for V2 in terms of V1

9.2*V1 - 6478 = 0.354*V1

V1(9.2-0.354) = 6478

Voltage across the current source

= V1+790*2.2 = 2470.32

V1 = 6478/(9.2-0.354) = VR4 = 732.32V

V2 = VR5 = 259.34V

With 3 significant digits:

Voltage across Current source: 2470V

VR4 =732V

VR5 = 259V

Assuming we can only justify 2 significant digits:

Voltage across current source = 2500V

VR4 = 730V

VR5 = 260V

Check by comparing power into ckt to power consumed:

2470.32V * 790mA = 1951.6W

0.79²*2200+732.32²/1e3 + (732.32-259.34)²/8.2e3 + 259.34²/7.19e3 = 1951.6W Checks