Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 months ago

# find voltage across resistor?

determine

the voltage across the current source?

voltage across R4?

voltage across R5?

Relevance
• 3 months ago

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

• 3 months ago

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