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(A) In the given circuit, the transistor is in a common-emitter configuration. The input circuit (base-emitter) is forward-biased.If the resistance $R

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The voltmeter shows a lower voltage.

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(A) In the given circuit, the transistor is in a common-emitter configuration. The input circuit (base-emitter) is forward-biased.If the resistance $R$ in the input circuit is increased, the base current $I_{B}$ decreases according to Ohm's law $(I_{B} = \frac{V_{in} - V_{BE}}{R})$.Since the collector current $I_{C}$ is related to the base current by the current gain $\beta$ $(I_{C} = \beta I_{B})$, a decrease in $I_{B}$ leads to a decrease in $I_{C}$.The bulb $B$ is in the collector circuit. The voltage across the bulb is given by $V_{B} = V_{CC} - I_{C}R_{bulb}$.As $I_{C}$ decreases, the voltage drop across the internal resistance of the transistor increases, and consequently, the voltage across the bulb $B$ decreases. Therefore, the voltmeter $V$ connected across the bulb will show a lower voltage.

In the given circuit, a voltmeter $V$ is connected across a bulb $B$. If the value of the resistor $R$ is increased, what happens?

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