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

If $V_{AB}=4V$ in the given figure, then resistance X will be:

1. 5 Ω

2. 10 Ω

3. 15 Ω

4. 20 Ω

The number of dry cells, each of e.m.f. 1.5 volt and internal resistance 0.5 ohm that must be joined in series with a resistance of 20 ohms, so as to send a current of 0.6 amperes through the circuit is 

(1) 2

(2) 8

(3) 10

(4) 12

For driving a current of 2 A for 6 minutes in a circuit, 1000 J of work is to be done. The e.m.f. of the source in the circuit is 

(1) 1.38 V

(2) 1.68 V

(3) 2.04 V

(4) 3.10 V

Two batteries of e.m.f. 4V and 8 V with internal resistances 1 Ω and 2 Ω are connected in a circuit with a resistance of 9 Ω as shown in figure. The current and potential difference between the points P and Q are 

(1) $\frac{1}{3}A\text{\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}}3V$

(2) $\frac{1}{6}A\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em} }4V$

(3) $\frac{1}{9}A\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em} 9}V$

(4) $\frac{1}{2}A\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}12V$

Four identical cells each having an electromotive force (e.m.f.) of 12V, are connected in parallel. The resultant electromotive force (e.m.f.) of the combination is :

(1) 48 V

(2) 12 V

(3) 4 V

(4) 3 V

The internal resistance of a cell of e.m.f. 12V is $5\times {10}^{-2}\Omega$. It is connected across an unknown resistance. The voltage across the cell, when a current of 60 A is drawn from it, is :

(1) 15 V

(2) 12 V

(3) 9 V

(4) 6 V

The current in the given circuit is 

(1) 0.1 A

(2) 0.2 A

(3) 0.3 A

(4) 0.4 A

Consider four circuits shown in the figure below. In which circuit power dissipated is greatest (Neglect the internal resistance of the power supply)

(1)

(2)

(3)

(4)

In the circuit, the reading of the ammeter is (assume internal resistance of the battery be zero) :

1. $\frac{40}{29}A$

2. $\frac{10}{9}A$

3. $\frac{5}{3}A$

4. 2 A