Exams › JEE Advanced › Physics

A capacitor of capacitance C discharges through a resistor R. Let t1 be the time for the stored energy to fall to half its initial value, and t2 be the time for the charge to fall to one-quarter of its initial value. What is the ratio t1/t2?

1. 2
2. 1
3. 1/2
4. 1/4

Correct answer: 1/4

Solution

Charge: Q(t) = Q0*e^(-t/RC). Energy: U(t) = Q²/(2C) = U0*e^(-2t/RC). For t1 (energy halved): e^(-2t1/RC) = 1/2 => t1 = RC*ln(2)/2. For t2 (charge reduced to Q0/4): e^(-t2/RC) = 1/4 => t2/RC = ln(4) = 2*ln(2) => t2 = 2RC*ln(2). Ratio: t1/t2 = [RC*ln(2)/2] / [2RC*ln(2)] = 1/4.

Related JEE Advanced Physics questions

• What is the final velocity attained by the loop?
• A long coaxial cable is made up of two cylindrical conductors with radii a and b. The inner conductor carries a constant current i, while the outer conductor acts as the return path for the current. What is the self-inductance of a segment of the cable of length l?
• Determine the inductance per unit length for a loop created by connecting the ends of two infinitely long parallel wires, each with radius r, and separated by a distance d between their centers. Ignore the magnetic flux inside the wires and the effects at the ends.
• If the electromotive force is 10 V and the rate of change of current is calculated as 2.5 A/s, what is the self-inductance of the coil, given by L = e / (dl/dt)?
• The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is
• A semicircular loop of radius r is positioned on the axis of a large current-carrying circular ring of radius R (with R >> r). The loop is placed at an axial distance of R*sqrt(3) from the center of the ring, and the plane of the semicircular loop makes an angle of 53 deg with the plane of the ring. Find the mutual inductance M between the two coils.

⚔️ Practice JEE Advanced Physics free + battle 1v1 →