Home Work # 1 (14-Feb) - Current Electricity - LIVE Short Duration REVISION Course on NEETprep LIVE AppContact Number: 8527521718Home Work # 1 (14-Feb) - Current Electricity - LIVE Short Duration REVISION Course on NEETprep LIVE AppContact Number: 8527521718The current through the \(5~\Omega\) resistor is:
| 1. | \(3.2~\text A\) | 2. | \(2.8~\text A\) |
| 3. | \(0.8~\text A\) | 4. | \(0.2~\text A\) |
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. \(\frac{20}{11}~\text{V}\)
What is the equivalent resistance of the circuit?
1. \(6~\Omega\)
2. \(7~\Omega\)
3. \(8~\Omega\)
4. \(9~\Omega\)
The total current supplied to the circuit by the battery is:
1. \(1~\text{A}\)
2. \(2~\text{A}\)
3. \(4~\text{A}\)
4. \(6~\text{A}\)
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
The potential difference \(V_{A}-V_{B}\) between the points \({A}\) and \({B}\) in the given figure is:
| 1. | \(-3~\text{V}\) | 2. | \(+3~\text{V}\) |
| 3. | \(+6~\text{V}\) | 4. | \(+9~\text{V}\) |
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. | \(\varepsilon_1-(i_1+i_2)R+i_1r_1=0\) |
| 1. | flow from \(A\) to \(B\) |
| 2. | flow in the direction which will be decided by the value of \(V\) |
| 3. | be zero |
| 4. | flow from \(B\) to \(A\) |
The current in a wire varies with time according to the equation \(I=(4+2t),\) where \(I\) is in ampere and \(t\) is in seconds. The quantity of charge which has passed through a cross-section of the wire during the time \(t=2\) s to \(t=6\) s will be:
| 1. | \(60\) C | 2. | \(24\) C |
| 3. | \(48\) C | 4. | \(30\) C |
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. | \(200\) |
If each resistance in the figure is of 9 Ω then reading of ammeter is
(1) 5 A
(2) 8 A
(3) 2 A
(4) 9 A
A potential divider is used to give outputs of \(2~\text{V}\) and \(3~\text{V}\) from a \(5~\text{V}\) source, as shown in the figure.
| 1. | \({R}_1=1~\text{k} \Omega, {R}_2=1 ~\text{k} \Omega, {R}_3=2 ~\text{k} \Omega\) |
| 2. | \({R}_1=2 ~\text{k} \Omega, {R}_2=1~\text{k} \Omega, {R}_3=2~\text{k} \Omega\) |
| 3. | \({R}_1=1 ~\text{k} \Omega, {R}_2=2~ \text{k} \Omega, {R}_3=2~ \text{k} \Omega\) |
| 4. | \({R}_1=3~\text{k} \Omega, {R}_2=2~\text{k} \Omega, {R}_3=2~ \text{k} \Omega\) |
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. | \(-{i}_2 {R}_2+{E}_2+{E}_3+{i}_3{R}_1=0\) |
In the circuit shown in the figure below, if the potential at point \(A\) is taken to be zero, the potential at point \(B\) will be:
1. \(+1\) V
2. \(-1\) V
3. \(+2\) V
4. \(-2\) V
For the network shown in the figure below, the value of the current \(i\) is:
1. \(\frac{18V}{5}\)
2. \(\frac{5V}{9}\)
3. \(\frac{9V}{35}\)
4. \(\frac{5V}{18}\)
