Exams › JEE Main › Physics

Two identical cells, each of emf E and internal resistance 1 ohm, are connected first in series and then in parallel across an external resistor R. Let J1 be the heating rate in R for the series combination and J2 for the parallel combination. If J1 = 2.25*J2, find R (in ohms).

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

Correct answer: 4

Solution

Setting [(2E/(R+2))/(E/(R+0.5))]² = 2.25 gives 2(R+0.5)/(R+2) = 1.5, leading to R = 4 ohm.

Related JEE Main Physics questions

• A physical quantity X is defined as ε0L(ΔV/Δt), where ε0 denotes the permittivity of free space, L is a length, ΔV is a potential difference, and Δt is a time interval. The dimensional formula of X matches that of which of the following?
• A set of n identical resistors is arranged once in a series combination and once in a parallel combination. What is the ratio of the larger equivalent resistance to the smaller equivalent resistance?
• A resistance coil is connected to a battery. In which arrangement will the maximum heat be produced?
• A bulb is rated at 220 V and 100 W. If the applied voltage falls by 2.5% from its rated value, by what percentage of the rated power will the power output decrease?
• An ammeter has a coil resistance of R. To extend its measuring range by a factor of n, the shunt resistance needed should be:
• How does the drift velocity Vd depend on the strength of the applied electric field?

⚔️ Practice JEE Main Physics free + battle 1v1 →