The voltage equation across the circuit is given by:
\[ V_A - iR - L \frac{di}{dt} - 12 = V_B \]
Rewriting, we get:
\[ V_A - V_B = iR + L \frac{di}{dt} + 12 \]
Given that the potential difference between points A and B is:
\[ V_A - V_B = +18 \, \text{volts} \]
\(V_A - V_B = +18 \, \text{volts}\)
The current passing through the battery in the given circuit, is:
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: