A current of 3A flows through the 2$\Omega$ resistor shown in the circuit. The power dissipated in the $5\Omega$ resistor is

 

1. 4 W                                             
2. 2 W
3. 1 W                                             
4. 5 W

A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively 1. .

A cell can be balanced against 110cm and 100 cm of potentiometer wire, respectively with and without being short-circuited through a resistance of $10 \Omega$. Its internal resistance is 

1. 1.0 $\Omega$

2. 0.5 $\Omega$

3. 2.0 $\Omega$

4. zero 

In the circuit shown, the current through the \(4~\Omega\) resistor is \(1~\text{A}\) when the points \(P\) and \(M\) are connected to a DC voltage source. The potential difference  between the points \(M\) and \(N\) is: 

1. \(1.5~\text{V}\)
2. \(1.0~\text{V}\)
3. \(0.5~\text{V}\)
4. \(3.2~\text{V}\)

In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs E₁ and E₂ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) $\frac{E_1}{E_2}$and  position of null point for the cell E₁.

How is the sensitivity of a potentiometer increased ?

Using Kirchoffs rules determine the value of unknown resistance R in the circuit so that no current flows through 4 Ω resistance. Also find the potential difference between A and D.

A capacitor of capacitance 5 μF is connected as shown in the figure. The internal resistance of the cell is 0.5 Ω. The amount of charge on the capacitor plate is?

1. 0 μC

2. 5 μC

3. 10 μC

4. 25 μC

In the circuit here, the steady state voltage across capacitor C is a fraction of the battery e.m.f. The fraction is decided by 

1. R₁ only

2. R₁ and R₂ only

3. R₁ and R₃ only

4. R₁, R₂ and R₃

In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state 

(1) $V_{AB}=V_{BC}=100V$

(2) $V_{AB}=75V,V_{BC}=25V$

(3) $V_{AB}=25V,V_{BC}=75V$

(4) $V_{AB}=V_{BC}=50V$

In the given figure each plate of capacitance C has partial value of charge equal to:

1. CE

2. $\frac{CE{R}_1}{R_2-r}$

3. $\frac{CE{R}_2}{R_2+r}$

4. $\frac{CE{R}_1}{R_1-r}$