Question

A transformer has a primary coil of 5200turns. When the output p.d is 12Vr.m.s, the input

p.d is 240Vr.m.s. (a) How many turns are there on the secondary coil? (b) The output

from the secondary coil is connected across a 20Ω lamp. (i)What is the r.m.s current in

the secondary coil? (ii) What is the r.m.s current in the primary coil?

(

Answer 1

Explanations & Calculations

• For an ideal transformer in which input power equals output power,

$\qquad\qquad \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$

• And most of the time transformers are assumed to be ideal.
• Secondary side is where the output is

a) Secondary turns ,

$\qquad\qquad \begin{aligned} \small \frac{N_s}{N_p} &= \small\frac{V_s}{V_p}\\ \small \frac{N_s}{5200} &=\small \frac{12V}{240}\\ \small N_s &= \small \bold{260\,\text{turns}} \end{aligned}$

b) 1)

Applying V =i*R to the lamp, the current flowing in the secondary coil could be found.

$\qquad\qquad \begin{aligned} \small i_s &= \small \frac{V}{R}\\ &= \small \frac{12V}{20\Omega}\\ &= \small \bold{0.6A = 600mA} \end{aligned}$

2) To find the primary current either relationship for an ideal transformer could be considered.

$\qquad\qquad \begin{aligned} \small \frac{i_p}{i_s} &= \small\frac{V_s}{V_p}\\ \small \frac{i_p}{0.6A} &= \small \frac{12V}{240V}\\ \small i_p &= \small \bold{0.03A = 30mA} \end{aligned}$