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Byju's Answer

Standard X

Physics

Step-Up and Step-Down Transformer

If the second...

Question

If the secondary winding of the ideal transformer shown in the circuit of figure has 40 turns, the number of turns in the primary winding for maximum power transfer to the 2 $Ω$ resistor will be

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Solution

The correct option is C 80
${(\frac{n_{2}}{n_{1}})}^{2} = \frac{| Z_{L} |}{| Z_{S} |}$

${(\frac{n_{2}}{n_{1}})}^{2} = \frac{2}{8} = \frac{1}{4}$

$\frac{n_{2}}{n_{1}} = \frac{1}{2}$

n $_{1}$ = 2n $_{2}$ = 2 x 40 = 80

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If the secondary winding of the ideal transformer shown in the circuit of figure has 40 turns, the number of turns in the primary winding for maximum power transfer to the 2 Ω resistor will be

The correct option is C 80(n2n1)2=|ZL||ZS| (n2n1)2=28=14 n2n1=12 n1 = 2n2 = 2 x 40 = 80

If the secondary winding of the ideal transformer shown in the circuit of figure has 40 turns, the number of turns in the primary winding for maximum power transfer to the 2 $Ω$ resistor will be

The correct option is C 80
${(\frac{n_{2}}{n_{1}})}^{2} = \frac{| Z_{L} |}{| Z_{S} |}$

${(\frac{n_{2}}{n_{1}})}^{2} = \frac{2}{8} = \frac{1}{4}$

$\frac{n_{2}}{n_{1}} = \frac{1}{2}$

n $_{1}$ = 2n $_{2}$ = 2 x 40 = 80