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Chapter 40: Problem 2930

In a transistor amplifier, \(\beta=62, \mathrm{R}_{\mathrm{L}}=5000 \Omega\) and internal resistance of the transistor is \(500 \Omega\). What will be the ratio of power amplification to voltage amplification? (A) 61 (B) 62 (C) 60 (D) 100

Short Answer

Expert verified
The ratio of power amplification to voltage amplification is 62. Therefore, the correct answer is (B) 62.

Step by step solution

01

Find the voltage gain

To find the voltage gain (A_v), we can use the formula: \( A_v = \dfrac{R_L}{R_L + r_i} \) Substituting the given values, we get: \( A_v = \dfrac{5000}{5000 + 500} \)
02

Calculate the voltage gain

Now, let's perform the computation: \( A_v = \dfrac{5000}{5500} = \dfrac{10}{11} \)
03

Find the power gain

The power amplification (A_p) is the product of the voltage gain (A_v) and the current gain (β). So, \( A_p = A_v \cdot \beta \) Substituting the values, we get: \( A_p = \dfrac{10}{11} \cdot 62 \)
04

Calculate the power gain

Let's compute the power gain: \( A_p = \dfrac{620}{11} \)
05

Find the ratio of power amplification to voltage amplification

Now, let's find the required ratio, which is the quotient of the power gain (A_p) and the voltage gain (A_v): \( \dfrac{A_p}{A_v} = \dfrac{\dfrac{620}{11}}{\dfrac{10}{11}} \)
06

Calculate the ratio

Lastly, compute the ratio: \( \dfrac{A_p}{A_v} = \dfrac{620}{10} = 62 \) The ratio of power amplification to voltage amplification is 62. Therefore, the correct answer is (B) 62.

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Most popular questions from this chapter

Which of the following the correct relationship between two current gains \(\alpha\) and \(\beta\) in a transistor? (A) \(\beta=[(1+\alpha) / \alpha]\) (B) \(\beta=[\alpha /(1+\alpha)]\) (C) \(\alpha=[(1+\beta) / \beta]\) (D) \(\alpha=[\beta /(1+\beta)]\)

What is the relation of the transistor current? (A) \(\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{E}}\) (B) \(\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{E}}+\mathrm{I}_{\mathrm{B}}\) (C) \(\mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{B}}\) (D) \(\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{E}}+\mathrm{I}_{\mathrm{B}}\)

In a common emitter amplifier, using output resistance of \(5000 \Omega\) and input resistance of \(2000 \Omega\), if the peak value of input signal voltage is \(10 \mathrm{mV}\) and \(\beta=50\) then what is the peak value of output voltage? (A) \(2.5 \times 10^{-4}\) Volt (B) \(5 \times 10^{-6}\) Volt (C) \(1.25\) Volt (D) 125 Volt

A transistor is connected in common emitter configuration. The collector supply is \(8 \mathrm{~V}\) and the voltage drop across a resistor of 800 in the collector circuit is \(0.5 \mathrm{~V}\). If the current gain factor \(\alpha\) is \(0.96\), then base current will be (A) \(27 \mu \mathrm{A}\) (B) \(26 \mu \mathrm{A}\) (C) \(25 \mu \mathrm{A}\) (D) \(24 \mu \mathrm{A}\)

The current gain of a transistor in common base mode is 0.99. To change the emitter current by \(5 \mathrm{~mA}\). What will be the necessary change in collector current ? (A) \(5.1 \mathrm{~mA}\) (B) \(4.95 \mathrm{~mA}\) (C) \(2.45 \mathrm{~mA}\) (D) \(0.196 \mathrm{~mA}\)
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