Question

Current I1, I2, I3 meet at a junction(node) in a circuit. All currents are marked as entering the node. If I1= -6sin(wt) mA and I2= 8cos(wt)mA, then I3 will be? ​

Answer 1

Given:

Current I1, I2, I3 meet at a junction(node) in a circuit. All currents are marked as entering the node. I1= -6sin(wt) mA and I2= 8cos(wt)mA

To find:

Variation of I3 ;

Calculation:

As per Kirchoff's Current Junction Law :

$\therefore \: \rm{I1 + I2 + I3 = 0}$

$= > \: \rm{ - 6 \sin( \omega t) + 8 \cos( \omega t) + I3 = 0}$

$= > \:I3 = \rm{ 6 \sin( \omega t) - 8 \cos( \omega t) }$

$\rm{= > \:I3 = (\sqrt{ {6}^{2} + {8}^{2} } ) \bigg \{ \frac{6}{ \sqrt{ {6}^{2} + {8}^{2} } } \sin( \omega t) - \frac{8}{ \sqrt{ {6}^{2} + {8}^{2} } } \cos( \omega t) } \bigg \}$

$\rm{= > \:I3 = (\sqrt{ 100 } ) \bigg \{ \frac{6}{ \sqrt{100 } } \sin( \omega t) - \frac{8}{ \sqrt{100 } } \cos( \omega t) } \bigg \}$

$\rm{= > \:I3 = (10 ) \bigg \{ \frac{6}{ 10} \sin( \omega t) - \frac{8}{10} \cos( \omega t) } \bigg \}$

Let 6/10 be $\cos(\theta)$

$\rm{= > \:I3 = (10 ) \bigg \{ \cos( \theta) \sin( \omega t) - \sin( \theta) \cos( \omega t) } \bigg \}$

$\rm{= > \:I3 = (10 ) \bigg \{ \sin( \omega t - \theta) } \bigg \}$

So, final answer is:

$\boxed{ \red{ \bold{= > \:I3 = (10 ) \bigg \{ \sin( \omega t - \theta) } \bigg \}}}$