Question

The current developed in a real conductor is given as $I=a{V}^{1/2}$, where V is the potential difference across the conductor. When it is connected with a resistance of $R\Omega$ in series with a battery of 6 Volt as shown in the figure, the power developed in the conductor is twice the power generated in resistance of $R\Omega$. Find the value of $R.(a=0.2AVol{t}^{-01/2})$

Answer 1

Apply KVL in loop

$iR+V=6$ (V= P.D across conductor)

$i=\frac{6-V}{R}\to \left(i\right)$

$a{V}^{\frac{1}{2}}=\frac{6-V}{R}\to \left(1\right)$

$P_{conductor}=VI$

$=V\times a{V}^{\frac{1}{2}}$

$=a{V}^{\frac{3}{2}}$

$P_{conductor}=i^{2}R$

$=\frac{(6-V{)}^2}{R^2}$

According to question

$P_{conductor}=2P_{resistance}$

$a{V}^{\frac{3}{2}}=\frac{2(6-V{)}^2}{R}\to \left(2\right)$

Divide (2) by (1)

$V^{\frac{3}{2}-\frac{1}{2}}=2(6-V)V=12-2V3V=12$

V = 4V

Put V in (2)

${aVV}^{\frac{1}{2}}=\frac{2(6-V{)}^2}{R}a\times 4\times 2=\frac{2\times 4}{R}\frac{1}{5}=\frac{1}{R}R=5\Omega$