Solution: Three Coils on a Toroid
Step 1: Determine Coupling Factor (Case 1)
When Coil B is open, no current flows through it ($I_B = 0$). The Voltmeter on Coil C is ideal ($I_C = 0$).
The setup acts like a simple transformer between A and C.
The voltage induced is proportional to the rate of change of current:
$$ V_A = L \frac{dI_A}{dt} = V_{\text{source}} $$
$$ V_C = M \frac{dI_A}{dt} $$
Given that $V_C = \frac{1}{2}V_{\text{source}}$, we divide the two equations:
$$ \frac{V_C}{V_A} = \frac{M}{L} = \frac{1}{2} \implies M = 0.5 L $$
Step 2: Setup Equations for Shorted Coil B (Case 2)
Now Coil B is short-circuited ($V_B = 0$). Current flows in both A and B.
The voltmeter on C reads the voltage induced by both A and B.
$$ \text{(A)} \quad V_{\text{source}} = L \dot{I}_A + M \dot{I}_B $$
$$ \text{(B)} \quad 0 = M \dot{I}_A + L \dot{I}_B $$
$$ \text{(C)} \quad V_{\text{reading}} = M \dot{I}_A + M \dot{I}_B $$
Step 3: Solve for Currents
From equation (B), we find the relationship between the currents:
$$ L \dot{I}_B = -M \dot{I}_A \implies \dot{I}_B = -\frac{M}{L} \dot{I}_A $$
Now substitute $\dot{I}_B$ into equation (A) to find $\dot{I}_A$:
$$ V_{\text{source}} = L \dot{I}_A + M(-\frac{M}{L}\dot{I}_A) = (L – \frac{M^2}{L}) \dot{I}_A $$
Step 4: Solve for Voltmeter Reading
Substitute $\dot{I}_B = -\frac{M}{L}\dot{I}_A$ into equation (C):
$$ V_{\text{reading}} = M(\dot{I}_A – \frac{M}{L}\dot{I}_A) = M \dot{I}_A (1 – \frac{M}{L}) $$
Now take the ratio of $V_{\text{reading}}$ to $V_{\text{source}}$:
$$ \frac{V_{\text{reading}}}{V_{\text{source}}} = \frac{M \dot{I}_A (1 – M/L)}{(L – M^2/L)\dot{I}_A} = \frac{M (1 – M/L)}{L(1 – M^2/L^2)} $$
Simplifying the algebra, this reduces to:
$$ \frac{V_{\text{reading}}}{V_{\text{source}}} = \frac{M}{L + M} $$
Substituting $M = 0.5L$:
$$ V_{\text{reading}} = V_{\text{source}} \times \frac{0.5L}{L + 0.5L} = \frac{V_{\text{source}}}{3} $$
Final Answer: The voltmeter reads V/3.
