In the given network $V_{1} = 100 ∠ 0^{0} V, V_{2} = 100 ∠ - 120^{0} V, V_{3} = 100 ∠ + 120^{0} V .$ The phasor current i (in Ampere) is

173.2 ∠ - 60^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

173.2 ∠ 120^{0}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

100.0 ∠ - 60^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100.0 ∠ 120^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $173.2 ∠ 120^{0}$
'i' is taken as output

Apply KCL at node

Node voltage is $100 ∠ 120^{0}$

$i = \frac{100 ∠ 120^{0} - 100 ∠ 0^{0}}{- j 1} + \frac{100 ∠ 120^{0} - 100 ∠ - 120^{0}}{j 1}$

$i = 173.2 ∠ - 60^{0} A$

Suggest Corrections

Similar questions

Q. A current

I

flows through a flat metal specimen of constant thickness and of the shape as shown in the figure. Given that

v_{1}, v_{2}

and

v_{3}

are mean drift speeds of the electron crossing the sections

P_{1}, P_{2}

and

P_{3}

respectively. Then

$V_{1}, V_{2} a n d V_{3}$ are the potential differences across the $1 Ω, 2 Ω a n d 3 Ω$ resistors and the flow of current is 5 A in a circuit .
Which of the following shows the correct values of $V_{1}, V_{2} a n d V_{3}$ measured in volts?

In the circuit shown in the figure, reading of voltmeter is $V_{1}$ when only $S_{1}$ is closed, $V_{2}$ when only $S_{2}$ is closed and $V_{3}$ when both $S_{1}$ and $S_{2}$ are closed. Choose the correct relation between $V_{1}$ , $V_{2}$ and $V_{3}$ .

A point moves with uniform acceleration and $v_{1}, v_{2}$ and $v_{3}$ denote the average velocities in the three successive intervals of time $t_{1}, t_{2}$ and $t_{3}$ . Which of the following relations is correct?

Q. The current i (in Ampere) in the 2

Ω

resistor of the given network is.

Join BYJU'S Learning Program

In the given network V1=100∠00V,V2=100∠−1200V,V3=100∠+1200V. The phasor current i (in Ampere) is

The correct option is B 173.2∠1200'i' is taken as output Apply KCL at node Node voltage is 100∠1200 i=100∠1200−100∠00−j1+100∠1200−100∠−1200j1 i=173.2∠−600A

In the given network $V_{1} = 100 ∠ 0^{0} V, V_{2} = 100 ∠ - 120^{0} V, V_{3} = 100 ∠ + 120^{0} V .$ The phasor current i (in Ampere) is

The correct option is B $173.2 ∠ 120^{0}$
'i' is taken as output

Apply KCL at node

Node voltage is $100 ∠ 120^{0}$

$i = \frac{100 ∠ 120^{0} - 100 ∠ 0^{0}}{- j 1} + \frac{100 ∠ 120^{0} - 100 ∠ - 120^{0}}{j 1}$

$i = 173.2 ∠ - 60^{0} A$