11.
In the circuits $$\left( a \right)$$ and $$\left( b \right)$$ switches $${S_1}$$ and $${S_2}$$ are closed at $$t = 0$$ and are kept closed for a long time. The variation of current in the two circuits for $$t \geqslant 0$$ are roughly shown by figure (figures are schematic and not drawn to scale) :
As we know, power $$P = {V_{rms}} \cdot {I_{rms}}\cos \phi $$
$${\text{as}}\,\cos \phi = 0\,\,\left( {\because \phi = {{90}^ \circ }} \right)$$
$$\therefore $$ Power consumed $$= 0$$ (in one complete cycle)
13.
An $$ac$$ voltage is applied to a resistance $$R$$ and an inductor $$L$$ in series. If $$R$$ and the inductive reactance are both equal to $$3\Omega ,$$ the phase difference between the applied voltage and the current in the circuit is
The phase difference $$\phi $$ is given by
$$\tan \phi = \frac{{{X_L}}}{R} = \frac{3}{3} = 1 \Rightarrow \phi = \frac{\pi }{4}.$$
14.
A sinusoidal voltage $$V\left( t \right) = 100\sin \left( {500t} \right)$$ is applied across a pure inductance of $$L = 0.02\,H.$$ The current through the coil is:
In a pure inductive circuit current always lags behind the emf by $$\frac{\pi }{2}.$$
If $$v\left( t \right) = {v_0}\sin \omega t\,{\text{then}}\,I = {I_0}\sin \left( {\omega t - \frac{\pi }{2}} \right)$$
Now, given $$v\left( t \right) = 100\sin \left( {500t} \right)$$
and $${I_0} = \frac{{{E_0}}}{{\omega L}} = \frac{{100}}{{500 \times 0.02}}\left[ {\because L = 0.02\,H} \right]$$
$${I_0} = 10\sin \left( {500t - \frac{\pi }{2}} \right) = - 10\cos \left( {500t} \right)$$
15.
The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $$\phi $$ linked with the primary coil is given by $$\phi = {\phi _0} + 4t,$$ where $$\phi $$ is in weber, $$t$$ is time in second and $${\phi _0}$$ is a constant, the output voltage across the secondary coil is
The magnetic flux linked with the primary coil is given by
$$\phi = {\phi _0} + 4t$$
So, voltage across primary
$$\eqalign{
& {V_p} = \frac{{d\phi }}{{dt}} = \frac{d}{{dt}}\left( {{\phi _0} + 4t} \right) \cr
& = 4\;V\,\,\left( {{\text{as }}{\phi _0} = {\text{constant}}} \right) \cr} $$
Also, we have $${N_p} = 50\,\,{\text{and}}\,\,{N_s} = 1500$$
As we know that voltage across primary and secondary coil is directly proportional to the no. of turns in primary and secondary coil respectively.
So, $$\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}}\,\,{\text{or}}\,\,{V_s} = {V_p}\frac{{{N_s}}}{{{N_p}}}$$
$$ = 4\left( {\frac{{1500}}{{50}}} \right) = 120\,V$$ NOTE
As in case of given transformer, voltage in secondary is increased, hence it is a step-up transformer.
16.
A circuit has a resistance of $$12\,ohm$$ and an impedance of $$15\,ohm.$$ The power factor of the circuit will be
17.
In a series $$LCR$$ circuit, the difference of the frequencies at which current amplitude falls to $$\frac{1}{{\sqrt 2 }}$$ of the current amplitude at resonance is
18.
The potential differences across the resistance, capacitance and inductance are $$80\,V,40\,V$$ and $$100\,V$$ respectively in an $$L-C-R$$ circuit. The power factor of this circuit is
19.
In a transformer, number of turns in the primary coil are 140
and that in the secondary coil are 280. If current in primary coil is $$4A,$$ then that in the secondary coil is
20.
In an electrical circuit $$R,L,C$$ and an $$a.c.$$ voltage source are all connected in series. When $$L$$ is removed from the circuit, the phase difference between the voltage the current in the circuit is $$\frac{\pi }{3}.$$ If instead, $$C$$ is removed from the circuit, the phase difference is again $$\frac{\pi }{3}.$$ The power factor of the circuit is :