A, B and C are voltmeters of resistance R, 1.5 R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and Vc respectively then,
VA =VB =Vc
The power dissipated in the circuit shiwn in the figure is 30 Watt. The value of R is
20 Ω
15Ω
10Ω
10Ω
A cell having an emf and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by
In an electrical circuit R, L, C and AC voltage source are all connected in series when L is removed from the circuit, the phase difference between the voltage and the current in the circuit is π/3. If instead, Cis removed from the circuit, the phase difference is again π/3. The power factor of the circuit is
1/2
1
1
C.
1
Here, phase difference
It is a condition of resonance, therefore, phase difference between voltage and current is zero and power factor is cos Φ = 1.
A millivoltmeter of 25 mV range is to be converted into an ammeter of 25 A range. The value (in ohm) of necessary shunt will be
0.001
0.01
1
1
A ring is made of wire having a resistance Ro = 12Ω. Find the points A and B, as shown in the figure, at which a current carrying conductor should be connected so that the resistance R of the subcircuit between these points is equal to 8/3Ω
The resistances in the two arms of the meter bridge are 5 ohm and R ohm, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 l1. The resistance R is,
10 ohm
15 ohm
20 ohm
20 ohm
A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of i) infinity ii) 9.5 ohm the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is,
0.25 ohm
0.95 ohm
0.5 ohm
0.5 ohm
In an ammeter 0.2 % of the main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be,
1/499 G
499/500 G
1/500 G
1/500 G
In the circuit shown the cells A and B have negligible resistances. For VA = 12V, R1 = 500Ω and R = 100Ω the galvanometer (G) shows no deflection. The value of VB is
4V
2V
12 V
12 V