A filament bulb (\(500\) W, \(100\) V) is to be used in a \(230\) V main supply. When a resistance\(R\) is connected in series, the bulb works perfectly and consumes \(500\) W. The value of \(R\) is:

1.	\(230\) $\mathrm{\Omega }$	2.	\(46\) $\mathrm{\Omega }$
3.	\(26\) $\mathrm{\Omega }$	4.	\(13\) $\mathrm{\Omega }$

Two metal wires of identical dimensions are connected in series. If \(\sigma_1~\text{and}~\sigma_2\) are the conductivities of the metal wires respectively, the effective conductivity of the combination is:

\(\mathrm{A, B}~\text{and}~\mathrm{C}\) are voltmeters of resistance \(R\), \(1.5R\) and \(3R\) respectively as shown in the figure above. When some potential difference is applied between \(\mathrm{X}\) and \(\mathrm{Y}\), the voltmeter readings are \({V}_\mathrm{A}\), \({V}_\mathrm{B}\) and \({V}_\mathrm{C}\) respectively. Then:

       

The figure given below shows a circuit when resistances in the two arms of the meter bridge are \(5~\Omega\) and \(R\), respectively. When the resistance \(R\) is shunted with equal resistance, the new balance point is at \(1.6l_1\). The resistance \(R\) is: 
           
1. \(10~\Omega\)
2. \(15~\Omega\)
3. \(20~\Omega\)
4. \(25~\Omega\)

If power dissipated in the \(9~\Omega\) resistor in the circuit shown is \(36\) W, the potential difference across the \(2~\Omega\) resistor will be:

            

A current of \(2~\text{A}\) flows through a \(2~\Omega\) resistor when connected across a battery. The same battery supplies a current of \(0.5~\text{A}\) when connected across a \(9~\Omega\) resistor. The internal resistance of the battery is: