The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common-emitter configuration is 1 k$\mathrm{\Omega }$. The peak value for an A.C input voltage of 0.01 V peak is:

1. 100 μA

2. 0.01 mA

3. 0.25 mA

4. 500 μA

When the NPN transistor is used as an amplifier:
1. the electrons move from base to collector.
2. the holes move from emitter to base.
3. the electrons move from collector to base.
4. the holes move from collector to emitter.

Use of transistor as an amplifier works in the:

1.  Cut off region

2.  Saturation region

3.  Passive region

4.  Active region

When a transistor is used as a switch, it is in:

1.  active state.

2.  cut-off state.

3.  saturation state.

4.  both cut-off and saturation states are possible.

In the circuit given below, if $V_{CE}$ is 5 V and $R=4\times {10}^5\Omega$, then current gain is:
(Neglect \(V_{BE}\))
           

1. 97

2. 98

3. 99

4. 100

When the emitter current increases by 10 mA, the base current increases by 0.4 mA. The value of voltage gain in the common emitter configuration of the amplifier will be:

(output and input resistance ratio is 50)

1. 1200

2. 1250

3. 150

4. 120

If the current gain in the transistor is 50, then $V_{CE}$ is equal to:
                  
1. 2 V

2. 3 V

3. 4 V

4. 6 V

A n-p-n transistor operates in a common emitter mode as shown. The value of \(R_{L}\) is:
\(\small\left ( \text{Given,}I_{C}=4 mA,~V_{CE}=4~V,V_{BE}=0.6~V~\text{and}~\beta _{dc}=100 \right )\\\)
        

1. 1 k$\Omega$

2. 2 k$\Omega$

3. 3 k$\Omega$

4. 4 k$\Omega$

The variation of output potential $V_{\mathrm{o}}$ with input potential $\left(V_i\right)$ in a transistor in CE-mode can be best represented as:

1.		2.
3.		4.

In an amplifier, an npn-transistor is used in common-emitter mode. If $\beta$ =25, then what will be the collector current if emitter current is 13 mA?

1. 13.5 mA

2. 12.5 mA

3. 11 mA

4. 10.5 mA