Question

Primary voltage is V_p, resistance of the primary winding is R_p. Turns in primary and secondary are respectively N_p

and N_s then secondary current in terms of primary voltage and secondary voltage respectively will be

• A. $\frac{V_{p}N_{p}}{R_{p}N_{s}},\frac{V_{s}N_{p}^{2}}{R_{p}N_{s}^{2}}$
• B. $\frac{V_{p}N_{p}^{2}}{R_{p}N_{s}},\frac{V_{s}^{2}N_{p}^{2}}{R_{p}N_{s}^{2}}$
• C. $\frac{V_{p}N_{p}}{R_{p}^{2}N_{s}},\frac{V_{s}N^{2}}{R_{p}^{2}N_{s}^{2}}$
• D. $\frac{V_{p}N_{p}^{2}}{R_{p}N_{s}^{2}},\frac{V_{s}^{2}N_{p}}{R_{p}^{2}N_{s}}$

Answer 1

Answer: (A)

$\frac{V_{p}N_{p}}{R_{p}N_{s}},\frac{V_{s}N_{p}^{2}}{R_{p}N_{s}^{2}}$

Answer 2

$\frac{i_{s}}{i_{p}} = \frac{N_{p}}{N_{s}}$

Now, according to the information given in the problem, i_p can be calculated by using the formula, V = iR so $i_{s} = \frac{V_{p}}{R_{p}} \times \frac{N_{p}}{N_{s}}$

(This is the secondary current in terms of V_p)

Now to rearrange the result obtained above, in terms of secondary voltage, we must replace the term of V_p in the above result by V_s. We know that $\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}$; $V_{p} = \frac{V_{s}N_{p}}{N_{s}}$, Substituting this in equation (i) $i_{s} = \frac{V_{s}}{R_{p}}\frac{N_{p}^{2}}{N_{s}^{2}}$

$\eta = \frac{P_{out}}{P_{in}} \times 100 = \frac{P_{out}}{V_{p}i_{p}} \times 100 = \frac{140}{240 \times 0.7} \times 100 = 83.3\%.$