If each resistance in the figure is \(9~\Omega\), then the reading of the ammeter is:

1. \(5~\text{A}\)
2. \(8~\text{A}\)
3. \(2~\text{A}\)
4. \(9~\text{A}\)

Equivalent resistance across terminals \(A\) and \(B\) will be:

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

The total current supplied to the circuit by the battery is:

1. \(1~\text{A}\)
2. \(2~\text{A}\)
3. \(4~\text{A}\)
4. \(6~\text{A}\)

In circuit shown below, the resistances are given in ohms and the battery is assumed ideal with emf equal to \(3\) volt. The voltage across the resistance \(R_4\) is:

1. \(0.4\) V

2. \(0.6\) V

3. \(1.2\) V

4. \(1.5\) V

A battery of emf \(E\) and internal resistance \(r\) is connected to a variable resistor \(R\) as shown below. Which one of the following is true​​​​​​?

The current in the arm CD of the circuit will be:
           
1. $i_1+i_2$

2. $i_2+i_3$

3. $i_1+i_3$

4. $i_1-i_2+i_3$

Consider the circuit shown in the figure below. The current \(I_3\) is equal to:

1. \(5\) A

2. \(3\) A

3. \(-3\) A

4. \(\frac{-5}{6}\) A

In the Wheatstone's bridge (shown in the figure below) \(X=Y\) and \(A>B\). The direction of the current between \(a\) and \(b\) will be:

In the circuit given below, the emf of the cell is \(2\) volt and the internal resistance is negligible. The resistance of the voltmeter is \(80\) ohm. The reading of the voltmeter will be:                                 
1. \(0.80\) volt
2. \(1.60\) volt
3. \(1.33\) volt
4. \(2.00\) volt

In the circuit shown below, \(E_1 = 4.0~\text{V}\), \(R_1 = 2~\Omega\), \(E_2 = 6.0~\text{V}\), \(R_2 = 4~\Omega\) and \(R_3 = 2~\Omega\). The current \(I_1\) is:

1. \(1.6\) A

2. \(1.8\) A

3. \(1.25\) A

4. \(1.0\) A