QUIZ 3:Current Electricity(#freetestseries) - JEE MCQ

 In the circuit shown, potential difference between X and Y will be 

 In the previous question, potential difference across the 40Ω resistance will be

 In the circuit shown, A and V are ideal ammeter and voltmeter respectively. Reading of the voltmeter will be 

When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell is [MP PMT 1996]

 The terminal potential difference of a cell when short-circuited is (E = E.M.F. of the cell)

 A primary cell has an e.m.f. of 1.5 volts, when short-circuited it gives a current of 3 amperes. The internal resistance of the cell is [CPMT 1976, 83]

A 50V battery is connected across a 10 ohm resistor. The current is 4.5 amperes. The internal resistance of the battery is [CPMT 1985; BHU 1997; Pb. PMT 2001]

The potential difference in open circuit for a cell is 2.2 volts. When a 4 ohm resistor is connected between its two electrodes the potential difference becomes 2 volts. The internal resistance of the cell will be [MP PMT 1984; SCRA 1994; CBSE PMT 2002]

A new flashlight cell of e.m.f. 1.5 volts gives a current of 15 amps, when connected directly to an ammeter of resistance 0.04Ω. The internal resistance of cell is [MP PET 1994]

A cell whose e.m.f. is 2 V and internal resistance is 0.1Ω, is connected with a resistance of3.9Ω. The voltage across the cell terminal will be [CPMT 1990; MP PET 1993; CBSE PMT 1999; AFMC 1999; Pb. PMT 2000; AIIMS 2001]

The reading of a high resistance voltmeter when a cell is connected across it is 2.2 V. When the terminals of the cell are also connected to a resistance of 5Ω the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell [KCET 2003; MP PMT 2003]

When cells are connected in parallel, then [MNR 1983]