Question

A battery of 10 volt is connected to a resistance of 20 ohm 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.

• A. 120 coulomb
• B. 120 log_e2 coulomb
• C. $\frac{120}{log_{e}2}$coulomb
• D. $\frac{60}{log_{e}2}$ coulomb

Answer 1

Answer: (B)

120 log_e2 coulomb

Answer 2

I = $\frac{dq}{dR}.\frac{dR}{dt} = \frac{V}{R}$ ; dq = 12 V $\frac{dR}{R}$

q = 12 V $\int_{20}^{40}\frac{dR}{R}$ = 12 V (log_e 40 – log_e20)

= 12 × 10 × log_e2, Hence correct answer is