In the given circuit,the power generated in t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the equivalent resistance of the parallel combination of $10 \, \Omega$ and $x \, \Omega$ be $R_p = \frac{10x}{10+x}$.The total resistance of

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let the equivalent resistance of the parallel combination of $10 \, \Omega$ and $x \, \Omega$ be $R_p = \frac{10x}{10+x}$.The total resistance of the circuit is $R_{eq} = \frac{1}{2} + R_p + 1 = \frac{3}{2} + \frac{10x}{10+x}$.The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{10}{\frac{3}{2} + \frac{10x}{10+x}}$.The current flowing through the $1 \, \Omega$ resistor is the total current $I$ of the circuit.The power generated in the $1 \, \Omega$ resistor is $P = I^2 R = I^2 (1) = I^2$.To maximize $P$,we must maximize $I$,which means we must minimize the total resistance $R_{eq}$.$R_{eq} = 1.5 + \frac{10x}{10+x}$.As $x$ increases,the term $\frac{10x}{10+x}$ increases. Therefore,to minimize $R_{eq}$,we must choose the smallest possible value for $x$.Given the options,the minimum value is $x = 0 \, \Omega$.

In the given circuit,the power generated in the $1 \, \Omega$ resistance will be maximum for $x$ equal to ................ $\Omega$.

Similar Questions

In the given circuit,the total power loss is $10 \text{ W}$. Find the value of resistance $R$ in $\Omega$.

For a cell, the graph between the potential difference $(V)$ across the terminals of the cell and the current $(I)$ drawn from the cell is shown in the figure. The $e.m.f.$ and the internal resistance of the cell are

If the ammeter in the given circuit reads $2\, A$,the resistance $R$ is ............ $\Omega$.

Twelve wires of equal length and equal cross-sectional area form a cube. If the resistance of each wire is $R$,then the effective resistance between the two ends of a diagonal is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module