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(A) For an ideal transformer, the ratio of voltages is equal to the ratio of the number of turns:$\frac{V_p}{V_s} = \frac{N_p}{N_s}$Given $V_p = 220 \

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(A) For an ideal transformer, the ratio of voltages is equal to the ratio of the number of turns:$\frac{V_p}{V_s} = \frac{N_p}{N_s}$Given $V_p = 220 \,V$, $N_p = 100$, and $N_s = 10$, we have:$\frac{220}{V_s} = \frac{100}{10} = 10$$V_s = \frac{220}{10} = 22 \,V$This secondary voltage $V_s = 22 \,V$ is applied across two resistors in series: $R_1 = 15 \,k\Omega$ and $R_2 = 7 \,k\Omega$.The output voltage $V_0$ is the voltage across the $7 \,k\Omega$ resistor, which can be calculated using the voltage divider rule:$V_0 = V_s 	imes \left( \frac{R_2}{R_1 + R_2} ight)$$V_0 = 22 	imes \left( \frac{7 \,k\Omega}{15 \,k\Omega + 7 \,k\Omega} ight)$$V_0 = 22 	imes \left( \frac{7}{22} ight) = 7 \,V$

The primary coil of a transformer is connected to a $220 \,V$ $AC$ source. The number of turns in the primary and secondary coils are $100$ and $10$ respectively. The secondary coil is connected to two resistors in series as shown in the figure. The output voltage $\left(V_0\right)$ is: (in $\,V$)

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