Step 1: Power Dissipation Formula
The power dissipation across a resistor is given by: \[ P = \frac{V^2}{R}. \]
Step 2: Resistance of Each Half
When the wire is cut into two equal halves, the resistance of each half becomes: \[ R_{\text{half}} = \frac{R}{2}. \]
Step 3: Power Dissipation Across Each Half
When \( V_0 \) is applied across each half: \[ P_{\text{half}} = \frac{V_0^2}{R_{\text{half}}} = \frac{V_0^2}{\frac{R}{2}} = \frac{2V_0^2}{R}. \]
Step 4: Total Power Dissipation Across Two Halves
The total power dissipation across the two halves is: \[ P_2 = 2 \times P_{\text{half}} = 2 \times \frac{2V_0^2}{R} = \frac{4V_0^2}{R}. \]
Step 5: Ratio of Power Dissipation
The original power dissipation \( P_1 \) is: \[ P_1 = \frac{V_0^2}{R}. \] The ratio \( P_2 : P_1 \) is: \[ \frac{P_2}{P_1} = \frac{\frac{4V_0^2}{R}}{\frac{V_0^2}{R}} = 4. \]
Step 6: Relation to \( \sqrt{x} \)
The ratio \( P_2 : P_1 \) is given as \( \sqrt{x} : 1 \), so: \[ \sqrt{x} = 4 \quad \Rightarrow \quad x = 16. \]
The current passing through the battery in the given circuit, is:
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ….. (Nearest integer).
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
Kirchhoffs Circuit Laws allow us to solve complex circuit problems.
It states that the “total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node“.
It states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero.