Three resistors \(1~\Omega , 2~\Omega\) and \(3~\Omega\) are combined in series. If the combination is connected to a battery of emf \(12~\text{V}\) and negligible resistance, the potential drop across \(2~\Omega\) resistor is:
1. \(2~\text{V}\)
2. \(5~\text{V}\)
3. \(4~\text{V}\)
4. \(6~\text{V}\)

Three resistors $\mathrm{}$\(2~\Omega, 4~\Omega\) and \(5~\Omega, \) are combined in parallel. If the combination is connected to a battery of emf \(20~\text V\) and negligible internal resistance, the total current drawn from the battery is:
1. \(10~\text A\)
2. \(17~\text A\)
3. \(13~\text A\)
4. \(19~\text A\)

At room temperature \((27~^\circ \text{C})\) the resistance of a heating element is \(100~\Omega.\) What is the temperature of the element if the resistance is found to be \(117~\Omega?\)
(Given that the temperature coefficient of the material of the resistor is \(1.70\times 10^{-4}~^{\circ}\text{C}^{-1}\))
1. \(1027~^{\circ}\text{C}\)
2. \(1007~^{\circ}\text{C}\)
3. \(1020~^{\circ}\text{C}\)
4. \(1127~^{\circ}\text{C}\)

A negligibly small current is passed through a wire of length \(15~\text{m}\) and uniform cross-section $\mathrm{}$\(6.0\times10^{-7}~\text{m}^2,\) and its resistance is measured to be $\mathrm{}$\(5.0~\Omega.\) What is the resistivity of the material at the temperature of the experiment?

A silver wire has a resistance of \(2.1~\Omega\) at \(27.5^\circ \text{C},\) and a resistance of \(2.7~\Omega\) at \(100^\circ \text{C}.\) The temperature coefficient of resistivity of silver is:
1. \(0.0033^\circ \text{C}^{-1}\)
2. \(0.039^\circ \text{C}^{-1}\)
3. \(0.0039^\circ \text{C}^{-1}\)
4. \(0.033^\circ \text{C}^{-1}\)

The current drawn from a \(12~\text{V}\) supply with internal resistance \(0.5~\Omega\) by the infinite network (shown in the figure) is:

    
1. \(3.12~\text{A}\)
2. \(3.72~\text{A}\)
3. \(2.29~\text{A}\)
4. \(2.37~\text{A}\)

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire. The value of ε is:

1. 1.33 V
2. 1.50 V
3. 1.24 V
4. 1.07 V

Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R=10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. The value of X is:

1. $12.1\Omega$

2. $12.4 \Omega$

3. $10.3 \Omega$

4. $11.7 \Omega$

The figure shows a 2.0 V potentiometer used for the determination of the internal resistance of a 1.5 V cell. The balance point of the cell in the open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. The internal resistance of the cell is:
    
1. \(1.68~\Omega \)
2. \(0.13~\Omega \)
3. \(0.31~\Omega \)
4. \(1.12~\Omega \)

A potential difference of 10 V is applied across a conductor of $1000 \mathrm{\Omega }$. The number of electrons flowing through the conductor in 300 sec is:

1. $1.875\times {10}^{16}$

2. $1.875\times {10}^{17}$

3. $1.875\times {10}^{22}$

4. $1.875\times {10}^{19}$