As in figure shown, if a capacitor C is charged by connecting it with resistance R, then energy  given by the battery will be 

(1) $\frac{1}{2}C{V}^2$

(2) More than $\frac{1}{2}C{V}^2$

(3) Less than $\frac{1}{2}C{V}^2$

(4) Zero

The resistance of a wire is \(R\) ohm. If it is melted and stretched to \(n\) times its original length, its new resistance will be:
1. \(nR\)
2. \(\frac{R}{n}\)
3. \(n^2R\)
4. \(\frac{R}{n^2}\)

A potentiometer is an accurate and versatile device to make electrical measurement of EMF because the method involves

1. cells

2. potential gradients 

3. a condition of no current flow through the galvanometer 

4. a combination of cells, galvanometer, and resistances 

The potential difference $\left({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}\right)$ between points A and B in the given figure is if the current is flowing from A to B:

(1) - 3 V                   

(2) +3 V

(3) +6 V                 

(4) +9 V

A potentiometer wire is 100 cm long and a constant potential is maintained across it. Two cells are connected in series first to support one another and then in the opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf is

1. 5:4              2. 3:4

3. 3:2              4. 5:1  

A potentiometer wire has a length 4 m and resistance 8Ω. The resistance that must be connected in series with the wire and an accumulator of emf 2V, so as to get a potential gradient 1mV per cm of the wire is 

(1)32Ω

(2)40Ω

(3)44Ω

(4)48Ω 

A, B, and C are voltmeters of resistance R, 1.5R, and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are V_A, V_B, and V_C respectively. Then,


1. V_A=V_B=V_C

2. V_A≠V_B=V_C

3. V_A=V_B≠V_C

4. V_A≠V_B≠V_C

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. $E_o$ and a resistance r₁. An unknown e.m.f. is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by

(1)$\frac{E_{o}rL}{\left(r_1\right)l}$

(2)$\frac{E_{o}rl}{\left(r+r_1\right)L}$

(3)E_ol/L

(4)$\frac{E_{o}rL}{\left(r+r_1\right)l}$

Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8V and the  average resistance per km is 0.5 Ω. The power loss in the wire is

(1) 19.2W

(2) 19.2kW

(3) 19.2J

(4) 12.2kW 

The resistances in the two arms of the meter bridge are 5 and R , respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6$l_1$. The resistance R, is


(1)10Ω

(2)15Ω

(3)20Ω

(4)25Ω