Question

A battery of 10 V is connected to a resistance of 20 $\Omega$ through a variable resistance R. The amount of charge which has passed in the circuit in 4 minutes, if the variable resistance R is increased at the rate
of 5 ohm min${}^{-1}$

• A. 120 C
• B. 120 ln2 C
• C. 240 ln2 C
• D. 60 ln2 C

Answer 1

The correct option is B 120 ln2 C

As $I=\frac{dq}{dt}=\frac{V}{R}$
$\Rightarrow \frac{dq}{dR}.\frac{dR}{dt}=\frac{V}{R}$
$\Rightarrow \frac{dq}{dR}.\frac{5}{60}=\frac{V}{R}$
$\Rightarrow q=12V{\int }_{20}^{40}\frac{dR}{R}\{\because R_i=20\omega ,R_f=20+5\times 4=40\Omega \}$
$\because R_i=20\Omega ,R_f=20+5\times =40\Omega )$
$\Rightarrow q=12\times 10V\times [lo{g}_{e}R{]}_{20}^{40}$
$q=120ln2Coulomb$
Alter:$\frac{dR}{dt}=\frac{5}{60}\frac{\Omega }{sec}$
$\Rightarrow R=\frac{t}{12}\Omega$
$\therefore I=\frac{10}{20+R}=\frac{10}{20+\frac{t}{12}}$
$\therefore \int dq={\int }_0^{4\times 60}=\frac{10}{20+R}\frac{10}{20+\frac{t}{12}}$
$q=10\times 12ln{\left[\frac{20+\frac{t}{12}}{20}\right]}_0^{4\times 60}=120ln2$