sue asked in Science & MathematicsPhysics · 2 months ago

# find voltage and currenth?

question 1: What is the voltage at Point A relative to the bottom rail?

question 2: What is the current through R1?

Relevance
• 2 months ago

Va=16.678v

IR1=29.782mA

The value I is current not currenth.

Figured with superposition.

Checked with spice.

• oubaas
Lv 7
2 months ago

using superposition :

I3 = 19.7/680-4.6*10^-3*560/680 = 0.025 A run by voltage source

Va = -I3*R3 = -3.0 V

I1 = (19.7+Va)/R1 = 16.7/560 = 0.030 A (30 mA)

• 2 months ago

Top of ckt = node V. Apply KCL above R3:

(V+19.7)/560 + V/120 = 0.0046

V(1/560+1/120) = 0.0046 - 19.7/560 = -0.030579A = 30.6mA

-0.030579/(1/560+1/120) = -3.02 <<<<< Q1

(V+19.7)/560 = (-3.02+19.7)/560 = 29.8mA  left into R1 and then DOWN into 19.7V  source <<<<<< Q2

Power from sources: (-3+4.6*1.5)*4.6 = 18mW

19.7*29.8 = 587mW -----> 587+18 = 0.6W

Consumed in resistance:

0.0298²*560+0.0046²*1500+0.0251²*120 = 0.6W checks

• 2 months ago

I'll use superpostion

1. voltage at A due to voltage source:

current source is open circuited.

I = 19.7/(560+120) = 0.0290 amp

Va = 120 x 0.0290 = 3.48 volts, negative

2. voltage due to current source. Voltage source is shorted.

120 parallel with 560 = 120•560/680 = 98.8 Ω

Va = 98.8 Ω x 4.6 ma = 455 mV, positive

add the two, Va = –3.48 + 0.455 = –3.02 volts

voltage across R1 = –19.7 –3.02 = –19.4 v

I = E/R = 19.4/560 = 0.0346 amp pointed left