The current flowing to the resistors ( 30Ω, 50Ω, 40Ω) is equals to 3/19 A. Therefore, the voltage drop across resistor 40Ω is... V = IR V = (3/19)(40) V = 120/19 V ≈ 6.3158 v
KCL lang po. Medyo mahaba-haba po yung solution pero same answer lang sa VDT. (1) kinuha ko yung total resistance ng circuit which is Rt = 27.1428 Ω (2) Total current ng circuit which is equal lang ng current sa 10Ω resistor = 21/19 or 1.1052 A (3) Apply KCL lang mula sa source (30V) hanggang sa node or junction ng 10, 20, at 30 Ω resistors. [ (30 - V,) / (10Ω) ] = 21/19 A V, = 18.9473 volts (4) KCL Current in resistors ( 30, 50, 40 Ω) = Total current - Current in 20Ω = (21/19) - ( 18/19) = 3/19 A = 0.1578 A Note* Current ng 30, 50, at 40 Ω ay pareha lang kasi naka-series silang tatlo.
6.31ohms?
Vab= 100||40/10+120||40(30)
V40=40/100×(100)(40)/100+40/10+(100)(40)/100+40(30)
V40=16/21V ≈ 0.7619V
Tama po ba?
6.32v
And 30 ohms= 4.74v
Is it,,,,,
6.32 V
The current flowing to the resistors ( 30Ω, 50Ω, 40Ω) is equals to 3/19 A.
Therefore, the voltage drop across resistor 40Ω is...
V = IR
V = (3/19)(40)
V = 120/19
V ≈ 6.3158 v
pano mo nakuha ang 3/19?
KCL lang po. Medyo mahaba-haba po yung solution pero same answer lang sa VDT.
(1)
kinuha ko yung total resistance ng circuit which is Rt = 27.1428 Ω
(2)
Total current ng circuit which is equal lang ng current sa 10Ω resistor = 21/19 or 1.1052 A
(3)
Apply KCL lang mula sa source (30V) hanggang sa node or junction ng 10, 20, at 30 Ω resistors.
[ (30 - V,) / (10Ω) ] = 21/19 A
V, = 18.9473 volts
(4)
KCL
Current in resistors ( 30, 50, 40 Ω) = Total current - Current in 20Ω
= (21/19) - ( 18/19)
= 3/19 A
= 0.1578 A
Note*
Current ng 30, 50, at 40 Ω ay pareha lang kasi naka-series silang tatlo.
@@klyde7042 so 18.95 is the V in 50 ohms?
No, 18.95v is yung voltage between sa node (V,)