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Figure 6 represents a step-down transformer with 500 turns in the primary and 50 turns in the secondary. The turns are wound uniformly on the core. The lengths of PQ and QR are indicated.

      

Figure 6 represents a step-down transformer with 500 turns in the primary and 50 turns in the secondary. The turns are wound uniformly on the core. The lengths of PQ and QR are indicated.
physics6520171214.jpg
Determine the p.d across PQ.

  

Answers


John
$\frac{V_p}{V_s}$ = $\frac{N_p}{N_s}$ = $V_s$

= $\frac{V_pN_s}{N_p}$ = $\frac{240 \times 50}{500}$ = 24 V

bt $V_{qp}$ = $\frac{1}{3}PR$ = V$_{qp}$ = $\frac{1}{3} \times 24 = 8V$
johnmulu answered the question on June 5, 2017 at 09:28


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