The logic behind the 'NOR' gate is that it gives:

1.	High output when both the inputs are low.
2.	Low output when both the inputs are low.
3.	High output when both the inputs are high.
4.	None of these

In a transistor circuit shown here, the base current is  35 $\mathrm{\mu }$A  and the potential difference across emitter-base junction is 4.5 V. The value of the resistor R_b is:
                 

1. 128.5 k$\mathrm{\Omega }$

2. 257 k$\mathrm{\Omega }$

3. 380.05 k$\mathrm{\Omega }$

4. None of these

In a transistor, a change of 8.0 mA in the emitter current produces a change of 7.8 mA in the collector current. What change in the base current is necessary to produce the same change in the collector current?
1. 50 $\mathrm{\mu }$A 
2. 100 $\mathrm{\mu }$A
3. 150 $\mathrm{\mu }$A 
4. 200 $\mathrm{\mu }$A

An NPN transistor conducts when:

A \(2\) V battery is connected across the points \(A\) and \(B\) as shown in the figure given below. Assuming that the resistance of each diode is zero in forward bias and infinity in reverse bias, the current supplied by the battery when its positive terminal is connected to \(A\) is:

In the circuit given below, if \(V(t)\) is the sinusoidal voltage source, then the voltage drop \(V_{AB}(t)\) across the resistance \(R\):

In a given circuit as shown the two input waveforms \(A\) and \(B\) are applied simultaneously. The resultant waveform \(Y\) is:
     

1.		2.
3.		4.

In a common emitter circuit, if V_CC is changed by 0.2 V, collector current changes by 4 x 10^–3 mA. The output resistance will be:

1. 10 k$\mathrm{\Omega }$ 

2. 30 k$\mathrm{\Omega }$ 

3. 50 k$\mathrm{\Omega }$ 

4. 70 k$\mathrm{\Omega }$

If the reverse bias in a junction diode is changed from \(5~\text V\) to \(15~\text V\) then the value of current changes from \(38~\mu \text{A}\) to \(88~\mu \text{A}.\) The resistance of the junction diode will be:
1. \(4\times10^{5}\) 
2. \(3\times10^{5}\)
3. \(2\times10^{5}\)
4. \(10^{6}\)