Exams › JEE Advanced › Physics

A flashlight bulb has resistance 1.5 ohm and is powered by 3 cells in series, each of emf 1.5 V and internal resistance 0.25 ohm. What is the potential difference across the bulb?

3 V
1.5 V
2.25 V
4.5 V

Correct answer: 3 V

Solution

Total emf 4.5 V across (1.5 + 0.75) ohm gives 2 A, so the bulb voltage is 2*1.5 = 3 V.

Related JEE Advanced Physics questions

The electric current in a conductor varies with time according to i = (20 + 4t). Determine the value of x if the total charge passing through a section of the conductor in 10 seconds is x × 10² C.
A 2.0 V potentiometer is used to find the internal resistance of a 1.5 V cell. With the cell's external circuit open, the balance length is 76.3 cm. When a resistor of 9.5 ohm is connected across the cell's terminals, the balance length drops to 64.8 cm. Determine the internal resistance of the cell (give the nearest integer value in ohm).
In a circuit, resistors of 2 ohm, 3 ohm and 4 ohm are connected in parallel across a battery of EMF 12 V and negligible internal resistance. Which resistor dissipates the maximum power?
An ideal cell of EMF 10 V and a non-ideal cell of EMF 15 V with internal resistance 1 ohm are connected in parallel between terminals A and B. What is the equivalent EMF between A and B?
In a Wheatstone bridge, resistors R1 and R2 occupy two adjacent arms while capacitors C1 and C2 occupy the other two adjacent arms. When a key is pressed the galvanometer shows no deflection. Which of the following expressions correctly represents the balance condition for this bridge?
In a potentiometer circuit used to measure an unknown resistance R, the balance (null) point is found at length PJ = 30 cm when a secondary cell of EMF Es = 10 V and internal resistance r = 1 ohm is used. When the secondary cell is replaced by another with Es = 5 V and r = 2 ohm, the balance point shifts to PJ = 10 cm. What is the value of R (in ohm)?

⚔️ Practice JEE Advanced Physics free + battle 1v1 →