The potential difference \(V_\mathrm{A}-V_\mathrm{B}\) between the points \(\mathrm{A}\) and \(\mathrm{B}\) in the given figure is:
1. | \(-3~\text{V}\) | 2. | \(+3~\text{V}\) |
3. | \(+6~\text{V}\) | 4. | \(+9~\text{V}\) |
In the following circuit, the battery \(E_1\) has an emf of \(12\) volts and zero internal resistance while the battery \(E\) has an emf of \(2\) volts. If the galvanometer \(G\) reads zero, then the value of the resistance \(X\) in ohms is:
1. | \(10\) | 2. | \(100\) |
3. | \(500\) | 4. | \(200\) |
Twelve wires of equal resistance \(R\) are connected to form a cube. The effective resistance between two diagonal ends \(A\) and \(E\) will be:
1. | \(\dfrac{5 R}{6}\) | 2. | \(\dfrac{6 R}{5}\) |
3. | \(12 R\) | 4. | \(3 R\) |
See the electrical circuit shown in this figure. Which of the following is a correct equation for it?
1. | \(\varepsilon_1-(i_1+i_2)R-i_1r_1=0\) |
2. | \(\varepsilon_2-i_2r_2-\varepsilon_1-i_1r_1=0\) |
3. | \(-\varepsilon_2-(i_1+i_2)R+i_2r_2=0\) |
4. | \(\varepsilon_1-(i_1+i_2)R+i_1r_1=0\) |
The current through the \(5~\Omega\) resistor is:
1. | \(3.2\) A | 2. | \(2.8\) A |
3. | \(0.8\) A | 4. | \(0.2\) A |
For the circuit given below, Kirchhoff's loop rule for the loop \(BCDEB\) is given by the equation:
1. | \(-{i}_2 {R}_2+{E}_2-{E}_3+{i}_3{R}_1=0\) |
2. | \({i}_2{R}_2+{E}_2-{E}_3-{i}_3 {R}_1=0\) |
3. | \({i}_2 {R}_2+{E}_2+{E}_3+{i}_3 {R}_1=0\) |
4. | \(-{i}_2 {R}_2+{E}_2+{E}_3+{i}_3{R}_1=0\) |
A battery of emf \(10\) V is connected to resistance as shown in the figure below. The potential difference \(V_{A} - V_{B}\)
between the points \(A\) and \(B\) is:
1. \(-2\) V
2. \(2\) V
3. \(5\) V
4. \(\frac{20}{11}~\text{V}\)
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