What is the equivalent resistance between the points A and D in given figure?

10 Ω
20 Ω
30 Ω
40 Ω

Solution

An ammeter has a resistance of G ohm and a range ofI amp.The value of resistance used in parallel to convert it into an ammeter of range nI amp is

nG
(n – 1)G
G/n
G/(n – 1)

Solution

A cell of internal resistance r is connected across an external resistance nr. Then the ratio of the terminal voltage to the emf of the cell is

¹⁄_n
¹⁄_n+1
ⁿ⁄_n+1
^n-1⁄_n

Solution

The resistance of the coil of an ammeter is R. The shunt required to increase its range n-fold should have a resistance

^R⁄_n
^R⁄_n-1
^R⁄_n+1
nR

Solution

In the equation AB = C, A is the current density, C is the electric field, Then B is

resistivity
resistivity
potential difference
resistance

Solution

J = σE ⇒ Jρ = E

J is current density, E is electric field

so B = ρ = resistivity.

A galvanometer of 50 ohm resistance has 25 divisions. A current of 4 × 10^–4 ampere gives a deflection of one division.To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of

2450 Ω in series.
2500 Ω in series.
245 Ω in series.
2550 Ω in series.

Solution

The deflection in a galvanometer decreases from 25 divisions to 5 divisions when a resistor of 20Ω is connected in series.Find resistance of galvanometer.

4 Ω
5 Ω
6 Ω
7 Ω

Solution

Potentiometer wire of length 1 m is connected in series with 490Ω resistance and 2 V battery. If 0.2 mV/cm is the potential gradient, then resistance of the potentiometer wire is

4.9 Ω
7.9 Ω
5.9 Ω
6.9 Ω

Solution

The four wires from a larger circuit intersect at junction A as shown. What is the magnitude and direction of the current between points A and B?

2 A from A to B
2 A from B to A
3 A from A to B
2 A from B to A

Solution

Kirchhoff’s junction rule states that the algebraic sum of all currents into and out of any branch point is zero:ΣI = 0. By convention, the sign of current entering a junction is positive and current leaving a junction is negative.

4A + 5 A – 6A + I_AB= 0, there fore I_AB= –3A. The wire between points A and B carries a current of 3A away from the junction.

A potentiometer consists of a wire of length 4m and resistance 10Ω. It is connected to a cell of e.m.f. 3V. The potential gradient of wire is

5V/m
2V/m
5V/m
10V/m

Solution

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