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Alexphysics edited Jun 25, 2026, 2:44 PM

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		+	### Statement
		+
		+	$11.3.18.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	In an ideal transformer (ideal means no losses, no leakage, and infinite magnetic permeability)
		+
		+	the voltage induced in each winding is proportional to the number of turns
		+
		+	$ V_1 = N_1 \frac{d\Phi}{dt}, V_2 = N_2 \frac{d\Phi}{dt}$
		+
		+	With the secondary winding short‑circuited, the voltage $ V_2 = 0$
		+	Since$ V_2 = N_2 \frac{d\Phi}{dt}$
		+	this implies that the magnetic flux $\Phi $in the core must be constant (or zero, if starting from zero initial conditions). For the net flux to be zero, the total magnetomotive force must be zero
		+
		+	$N_1 I_1 + N_2 I_2 = 0$
		+
		+	From this, the ratio of magnitudes is :
		+
		+	$\boxed{\frac{I_1}{I_2} = \frac{N_2}{N_1}}$
		+
		+	The negative sign indicates that the currents are in opposite phase, but the ratio of amplitudes is as given.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$11.3.18.$ [Insert the problem statement]

			### Solution

			In an ideal transformer (ideal means no losses, no leakage, and infinite magnetic permeability)

			the voltage induced in each winding is proportional to the number of turns

			$ V_1 = N_1 \frac{d\Phi}{dt}, V_2 = N_2 \frac{d\Phi}{dt}$

			With the secondary winding short‑circuited, the voltage $ V_2 = 0$
			Since$ V_2 = N_2 \frac{d\Phi}{dt}$
			this implies that the magnetic flux $\Phi $in the core must be constant (or zero, if starting from zero initial conditions). For the net flux to be zero, the total magnetomotive force must be zero

			$N_1 I_1 + N_2 I_2 = 0$

			From this, the ratio of magnitudes is :

			$\boxed{\frac{I_1}{I_2} = \frac{N_2}{N_1}}$

			The negative sign indicates that the currents are in opposite phase, but the ratio of amplitudes is as given.

			#### Answer

			[Insert a concise answer or boxed result]