What is total resistance across terminals \(A\) and \(B\) in the following network?

1.	\(R\)	2.	\(2R\)
3.	\(\dfrac{3R}{5}\)	4.	\(\dfrac{2R}{3}\)

The Wheatstone bridge shown in the figure below is balanced when the uniform slide wire \(AB\) is divided as shown. Value of the resistance \(X\) is:
          

1. \(3~\Omega\)
2. \(4~\Omega\)
3. \(2~\Omega\)
4. \(7~\Omega\)

The current in a wire varies with time according to the equation \(I=(4+2t),\) where \(I\) is in ampere and \(t\) is in seconds. The quantity of charge which has passed through a cross-section of the wire during the time \(t=2\) s to \(t=6\) s will be:

An ammeter A of finite resistance and a resistor R are joined in series to an ideal cell C. A potentiometer P is joined in parallel to R. The ammeter reading is ${\mathrm{I}}_0$ and the potentiometer reading is V₀ . P is now replaced by a voltmeter of finite resistance. The ammeter reading now is I and the voltmeter reading is V.

It can be concluded that:

1. $\mathrm{I}>{\mathrm{I}}_0,$ $\mathrm{V}<{\mathrm{V}}_0$

2. $\mathrm{I}>{\mathrm{I}}_0,$ $\mathrm{V}={\mathrm{V}}_0$

3. $\mathrm{I}={\mathrm{I}}_0,$ $\mathrm{V}<{\mathrm{V}}_0$

4. $\mathrm{I}<{\mathrm{I}}_0,$ $\mathrm{V}={\mathrm{V}}_0$

A voltmeter of resistance \(660~\Omega\)$\mathrm{}$ reads the voltage of a very old cell to be \(1.32\) V while a potentiometer reads its voltage to be \(1.44\) V. The internal resistance of the cell is:
1. \(30~\Omega\)$\mathrm{}$ 
2. \(60~\Omega\)
3. \(6~\Omega\)
4. \(0.6~\Omega\)$\mathrm{}$

The power dissipated across the \(8~\Omega\) resistor in the circuit shown here is \(2~\text{W}\). The power dissipated in watts across the \(3~\Omega\) resistor is:
       

Three resistances \(\mathrm P\), \(\mathrm Q\), and \(\mathrm R\), each of \(2~\Omega\) and an unknown resistance \(\mathrm{S}\) form the four arms of a Wheatstone bridge circuit. When the resistance of \(6~\Omega\) is connected in parallel to \(\mathrm{S}\), the bridge gets balanced. What is the value of \(\mathrm{S}\)?