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

A transistor circuit in a common emitter configuration is shown. The current gain is 100. If \(V_{BE}\approx 0,\) then \(V_{CE}\) is given by:

1. 6 V

2. 5 V

3. 4 V

4. 3 V

An $n-p-n$ transistor having A.C. current gain of 50 is to be used to make an amplifier of voltage gain 6. What will be the power gain in the amplifier?

1. 300

2. 600

3. 750

4. 400

In an NPN transistor, 10⁸ electrons enter the emitter in ${10}^{-8}$s. If 1% electrons are lost in the base, the fraction of current that enters the collector and the current amplification factor are, respectively:

1. 0.7 and 50

2. 0.9 and 90

3. 0.8 and 49

4. 0.99 and 99

When a transistor is used in a common emitter mode as an amplifier: