Question

Two electric devices A and B are connected in parallel, and the rms current in A is 15 A. If the current in B lags A by n/3 radians and the line current is 23.4 A, determine the current in B.

Answer 1

Concept

Electronic devices are components for controlling the flow of electrical currents for the aim of knowledge processing and system control.

Given

Given that two devices are connected in parallel and also the rms current in $A=15Amp$ and current in $B$ lags $A$ by $\frac{\pi}{3}$.

Find

We have to look out the current in $B.$

Solution

RMS current of device $A$ is $15Amp$.

Rms current of device $B$ is lags behind $A$ by $\frac{\pi}{3}$.

Let magnitude of current of device $B$ is $x.$

Therefore, RMS current of device $B$ is $xL\frac{\pi}{3}$ or $xL60^{\circ}$.

Now, we'll find the overall current line

Total line current $=$ Current of device $A+$ current of device $B$

$\begin{aligned}23.4&=15+xL60^{\circ}\\ 8.4&=xL60^{\circ}\end$

Hence, the RMS current of device $B$ is $8.4L60^{\circ}$.

#SPJ1

Answer 2

We know that electronic devices, in general, are those devices that are responsible for controlling the electric current flow in order to control specific systems.

Given:

We have been given that there are two devices connected to each other in parallel combination.

• RMS value of the current = $A = 15\ A$
• Current in $B$ lags by = $A =$ $\frac{ \pi}{3}$

To Find:

• Current in $B$

Solution:

• We know that the RMS value of $A$ is $15 \ A$.
• We also know that the RMS value of $B$ lags behind the device $A$ by $\frac{ \pi}{3}$.

Therefore, the RMS current value of the device $B$ = $xL\frac{\pi}{3}$ = $xL60$°

(∵ $\frac{\pi}{3} =$ $60$°)

To find the current in the device $B$, we will use the following formula:

$I_{Total} =\sqrt{I^{2} _{A} +I^{2}_{B} -2I_{A}I_{B} (cos 180- \theta) }$

Putting the given values in the above equation we get,

$23.4 \ A = \sqrt{15^{2} + I^{2}_{B} -2 \times 15 \times I_{B} (cos 180 - \theta) }$

$I_{B} = 11.96 \ A$

Hence, the RMS current of the device $B$ is $11.96 \ A$

#SPJ1