Question

in a transformer without any loose in power,there are 6000 turns in the primary and 240 in the secondary.The primary voltage is 100 V and the current is 0.2 A. find the voltage and current in the secondary​

Answer 1

Given :-

• Number of turns in primary, $\sf N_p = 6000$

• Number of turns in secondary, $\sf N_s = 240$

• Primary voltage, $\sf V_p = 240V$

• Current in primary, $\sf I_p = 0.2A$

To Find :-

• Secondary voltage, $\sf V_s$

• Current in secondary, $\sf I_s$

Solution :-

We know :-

$\underline{\boxed{\bf \dfrac{V_s}{V_p}= \dfrac{N_s}{N_p}}}$

$\longrightarrow \sf V_s = \dfrac{N_s \times V_p}{N_p} \\\\\longrightarrow V_s = \dfrac{240 \times 100}{6000}\\\\\longrightarrow V_s = \dfrac{24}{6} \\\\\longrightarrow \bf V_s = 4V$

Secondary voltage, $\bf V_s = 4V$

We know :-

$\underline{\boxed{\bf \dfrac{I_p}{I_s}= \dfrac{V_s}{V_p}}}$

$\longrightarrow \sf I_p \times V_p = V_s \times I_s \\\\\longrightarrow I_s = \dfrac{ I_p \times V_p }{V_s} \\\\\longrightarrow I_s = \dfrac{0.2 \times 100}{4} \\\\\longrightarrow I_s = \dfrac{20}{4} \\\\\longrightarrow \bf I_s = 5A$

Current in secondary = 5A

Answer 2

Given :-

Number of turns in primary, \sf N_p = 6000N _p

=6000

Number of turns in secondary, \sf N_s = 240N _s

=240

Primary voltage, V_p = 240VV _p

=240V

Current in primary, I_p = 0.2AI _p

=0.2A

To Find :-

Secondary voltage, V_s

Current in secondary, I_s

Solution :-

We know :-

V _p×V _s

= N _p×N _s

⟶V _s = N _s/N _s ×V _p

⟶V _s

= 6000/240×100

⟶V _s = 6/24

⟶V _s =4V

Secondary voltage

V_s = 4VV_s =4V

We know :-

I _s×I _p

= V _p×V _s

⟶I _p ×V _p =V _s ×I _s

⟶I _s = V _s×I _p ×V _p

⟶I _s = 4/0.2×100

⟶I _s = 4/20

⟶I _s =5A

Current in secondary = 5A