EMI BYU 36

Solution 36: Capacitor Bridge (User Method)

36. Capacitor Bridge (Energy Method)

Method: Work-Energy Theorem ($W_{\text{battery}} = \Delta U_{\text{cap}} + K_{\text{ind}}$).
We compare the equivalent capacitance of the system in the Initial State vs. the Max Current State.

Step 1: Initial Equivalent Capacitance ($S$ Open)

Initially, the bridge is open. The left branch and right branch are independent series combinations connected to the battery.

  • Left Branch ($C, C$ series): $C_L = C/2$.
  • Right Branch ($C, 3C$ series): $C_R = \frac{C \cdot 3C}{C + 3C} = \frac{3C}{4}$.
$$ C_{eq, i} = \frac{C}{2} + \frac{3C}{4} = \frac{5C}{4} $$ $$ U_i = \frac{1}{2} C_{eq, i} \mathcal{E}^2 = \frac{5}{8} C \mathcal{E}^2 $$

Step 2: Max Current State ($V_L = 0$)

When current in the inductor is maximum, the induced EMF is zero ($L di/dt = 0$). This means the potential difference across the inductor is zero. This effectively shorts the middle nodes, changing the topology:

  • Top Capacitors ($C, C$) become parallel: $C_{top} = 2C$.
  • Bottom Capacitors ($C, 3C$) become parallel: $C_{bot} = 4C$.
  • These two groups are now in series with the battery.
$$ C_{eq, f} = \frac{C_{top} C_{bot}}{C_{top} + C_{bot}} = \frac{(2C)(4C)}{2C + 4C} = \frac{8C^2}{6C} = \frac{4C}{3} $$ $$ U_f = \frac{1}{2} C_{eq, f} \mathcal{E}^2 = \frac{4}{6} C \mathcal{E}^2 = \frac{2}{3} C \mathcal{E}^2 $$

Step 3: Work Done and Energy Balance

The battery does work to supply the extra charge required by the increase in capacitance.

$$ \Delta Q = (C_{eq, f} – C_{eq, i}) \mathcal{E} = \left( \frac{4C}{3} – \frac{5C}{4} \right) \mathcal{E} = \left( \frac{16-15}{12} \right) C \mathcal{E} = \frac{C \mathcal{E}}{12} $$ $$ W_{\text{bat}} = \Delta Q \cdot \mathcal{E} = \frac{C \mathcal{E}^2}{12} $$

Using Conservation of Energy ($W_{\text{bat}} = \Delta U_{\text{cap}} + U_{\text{ind}}$):

$$ \frac{C \mathcal{E}^2}{12} = (U_f – U_i) + \frac{1}{2} L i^2 $$ $$ \frac{C \mathcal{E}^2}{12} = \left( \frac{2}{3} – \frac{5}{8} \right) C \mathcal{E}^2 + \frac{1}{2} L i^2 $$ $$ \frac{C \mathcal{E}^2}{12} = \left( \frac{16 – 15}{24} \right) C \mathcal{E}^2 + \frac{1}{2} L i^2 $$ $$ \frac{2 C \mathcal{E}^2}{24} – \frac{1 C \mathcal{E}^2}{24} = \frac{1}{2} L i^2 $$ $$ \frac{C \mathcal{E}^2}{24} = \frac{1}{2} L i^2 \implies i^2 = \frac{C \mathcal{E}^2}{12 L} $$
$$ i_{max} = \mathcal{E} \sqrt{\frac{C}{12 L}} $$
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