Q. No.The figure shows a circuit that contains three identical resistors with resistance \(R = 9.0~\Omega\) each, two identical inductors with inductance \(L = 2.0~\text{mH}\) each, and an ideal battery with emf \(\varepsilon = 18~\text{V}\). The current \('i'\) through the battery just after the switch is closed will be:
1. \(0.2~\text{A}\)
2. \(2~\text{A}\)
3. \(4~\text{A}\)
4. \(2~\text{mA}\)
1. \(0.2~\text{A}\)
2. \(2~\text{A}\)
3. \(4~\text{A}\)
4. \(2~\text{mA}\)
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. | \(+9~\text{V}\) |
A filament bulb \((500~\text{W},100~\text{V})\) is to be used in a \(230~\text{V}\) main supply. When a resistance\(R\) is connected in series, the bulb works perfectly and consumes \(500~\text{W}.\) The value of \(R\) is:
| 1. | \(230~\Omega\) | 2. | \(46~\Omega\) |
| 3. | \(26~\Omega\) | 4. | \(13~\Omega\) |
2. 3 : 4
3. 3 : 2
4. 5 : 1
| 1. | \(\dfrac{a^3R}{3b}\) | 2. | \(\dfrac{a^3R}{2b}\) |
| 3. | \(\dfrac{a^3R}{b}\) | 4. | \(\dfrac{a^3R}{6b}\) |
Two metal wires of identical dimensions are connected in series. If \(\sigma_1~\text{and}~\sigma_2\)
| 1. | \(\frac{2\sigma_1 \sigma_2}{\sigma_1+\sigma_2}\) | 2. | \(\frac{\sigma_1 +\sigma_2}{2\sigma_1\sigma_2}\) |
| 3. | \(\frac{\sigma_1 +\sigma_2}{\sigma_1\sigma_2}\) | 4. | \(\frac{\sigma_1 \sigma_2}{\sigma_1+\sigma_2}\) |
A circuit contains an ammeter, a battery of \(30~\text{V},\) and a resistance \(40.8~\Omega\) all connected in series. If the ammeter has a coil of resistance \(480~\Omega\) and a shunt of \(20~\Omega,\) then the reading in the ammeter will be:
1. \(0.5~\text{A}\)
2. \(0.02~\text{A}\)
3. \(2~\text{A}\)
4. \(1~\text{A}\)
A potentiometer wire of length \(L\) and a resistance \(r\) are connected in series with a battery of EMF \(E_{0 }\) and resistance \(r_{1}\). An unknown EMF is balanced at a length l of the potentiometer wire. The EMF \(E\) will be given by:
1. \(\frac{L E_{0} r}{l r_{1}}\)
2. \(\frac{E_{0} r}{\left(\right. r + r_{1} \left.\right)} \cdot \frac{l}{L}\)
3. \(\frac{E_{0} l}{L}\)
4. \(\frac{L E_{0} r}{\left(\right. r + r_{1} \left.\right) l}\)
A potentiometer wire has a length of \(4~\text{m}\) and resistance \(8~\Omega.\) The resistance that must be connected in series with the wire and an energy source of emf \(2~\text{V}\), so as to get a potential gradient of \(1~\text{mV}\) per cm on the wire is:
1. \(32~\Omega\)
2. \(40~\Omega\)
3. \(44~\Omega\)
4. \(48~\Omega\)
\({A, B}~\text{and}~{C}\) are voltmeters of resistance \(R,\) \(1.5R\) and \(3R\) respectively as shown in the figure above. When some potential difference is applied between \({X}\) and \({Y},\) the voltmeter readings are \({V}_{A},\) \({V}_{B}\) and \({V}_{C}\) respectively. Then:
| 1. | \({V}_{A} ={V}_{B}={V}_{C}\) | 2. | \({V}_{A} \neq{V}_{B}={V}_{C}\) |
| 3. | \({V}_{A} ={V}_{B}\neq{V}_{C}\) | 4. | \({V}_{A} \ne{V}_{B}\ne{V}_{C}\) |
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