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+### Statement
+
+$11.3.24.$ [Insert the problem statement]
+
+### Solution
+
+When an alternating voltage is applied to a coil, the total flux generated is split into two identical paths, so that the other coil, placed in one of the branches, is only linked by half of the total flux.
+
+In the first case, with$ V_1 = 40\ \text{V} $on coil 1, a voltage$ V_2 = 10\ \text{V} $appears on coil 2. The voltage ratio is
+
+$\frac{V_2}{V_1} = \frac{N_2}{2N_1} = \frac{1}{4}$
+
+from which it follows that $N_2 = N_1/2$ (coil 2 has half the number of turns as coil 1).
+
+When the connection is reversed and $V_2' = 10\ \text{V}$ is applied to coil 2, the generated flux divides just as before. Coil 1 is in one branch and receives half of the total flux. The voltage induced in it is
+
+$V_1' = N_1 \frac{d}{dt}\!\left(\frac{\Phi_2}{2}\right) = \frac{N_1}{2N_2}\,V_2'$
+
+Substituting $N_2 = N_1/2 $ gives
+
+$V_1' = V_2' = 10\ \text{V}$
+
+#### Answer
+
+[Insert a concise answer or boxed result]