If the lightbulb A in the circuit shown in the image burned out, the path for the current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.
If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.
Answer:0.1759 v
Explanation:
Intensity of wave at receiver end is
I=
I=
I=
Amplitude of electric field at receiver end
Amplitude of induced emf
=
=
=
In linear motion , when a body moves with uniform velocity , in time t , its linear displacement will be ;
S = r∅ S = vt
r∅ = vt
r.∅ / t = v
As
v = rw
where ∅ = 90° is the angle between between radius vector r and angular velocity w (omega )
In case ∅ ≠ 90° , we can write v = r w sin∅
It gives us v = w× r
