In the figure shown `(sin i)/(sin r)` is equal to:

A ray of light travels from a medium of refractive index n_(1) to a medium of refractive index n_(2) . If angle of incidence is i and angle of refraction is r, than (sin i)/(sin r) is equal to

Consider the circuit shown in the figure. The current i_(3) is equal to

sum_(r=0)^(n)nC_(r)(sin rx) is equal to

If I _(n)=int _(0)^(pi) (sin (2nx))/(sin 2x)dx, then the value of I _( n +(1)/(2)) is equal to (n in I) :

If I _(n)=int _(0)^(pi) (sin (2nx))/(sin 2x)dx, then the value of I _( n +(1)/(2)) is equal to (n in I) :

The power dissipated in the resistance R, in the circuit as shown in the figure, is maximum if R is equal to

Consider the circuit shown in the figure. The current I_(3) is equal to

Consider the circuit shown in the figure. The current I_(3) is equal to