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

Solution :
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.