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. | \(\varepsilon_1-(i_1+i_2)R+i_1r_1=0\) |
A current of \(3~\text{A}\) flows through the \(2~\Omega\) resistor shown in the circuit. The power dissipated in the \(5~\Omega\) resistor is:
1. | \(4~\text{W}\) | 2. | \(2~\text{W}\) |
3. | \(1~\text{W}\) | 4. | \(5~\text{W}\) |
1. | \(6.3\) min | 2. | \(8.4\) min |
3. | \(12.6\) min | 4. | \(4.2\) min |
The total power dissipated in watts in the circuit shown below is:
1. | \(16\) W | 2. | \(40\) W |
3. | \(54\) W | 4. | \(4\) W |
Three resistances \(\mathrm P\), \(\mathrm Q\), and \(\mathrm R\), each of \(2~\Omega\) and an unknown resistance \(\mathrm{S}\) form the four arms of a Wheatstone bridge circuit. When the resistance of \(6~\Omega\) is connected in parallel to \(\mathrm{S}\), the bridge gets balanced. What is the value of \(\mathrm{S}\)?
1. | \(2~\Omega\) | 2. | \(3~\Omega\) |
3. | \(6~\Omega\) | 4. | \(1~\Omega\) |
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 power dissipated across the \(8~\Omega\) resistor in the circuit shown here is \(2~\text{W}\). The power dissipated in watts across the \(3~\Omega\) resistor is:
1. | \(2.0\) | 2. | \(1.0\) |
3. | \(0.5\) | 4. | \(3.0\) |
1. | \(2:1\) | 2. | \(4:9\) |
3. | \(9:4\) | 4. | \(1:2\) |
1. | \(26.7\) cm | 2. | \(33.4\) cm |
3. | \(46.7\) cm | 4. | \(96.7\) cm |
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 |