At resonance, impedance is given by:
\[ Z = R. \]
The current in the circuit is:
\[ I = \frac{V}{R}. \]
When the resistance \(R\) is halved:
\[ R \to \frac{R}{2}, \quad I \to 2I. \]
Thus, the current amplitude becomes double.
Final Answer: Double.
A | B | Y |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is: