91Ó°ÊÓ

In a p-n-p transistor, the emitter current is essentially the total current flowing out of the emitter. The worked problem showed us that it's important to understand the relationship between the different currents in a transistor. For a p-n-p transistor, the emitter current \( I_e \) is given by the sum of the collector current \( I_c \) and the base current \( I_b \). Mathematically, it's represented as:\[I_e = I_c + I_b\]In this particular problem, we are asked to find the emitter current given that 90% of holes emitted by the emitter reach the collector. This means that the efficient part of the emitter current goes to provide the collector current and this efficiency is described by:\[I_c = 0.9 \, I_e\]To find the emitter current, we rearrange the formula to solve for \( I_e \), where the collector current \( I_c \) is known to be \( 10 \, \text{mA} \). Plugging into the equation provides:\[I_e = \frac{10}{0.9} \approx 11.11 \, \text{mA}\]Understanding that this value needs to reflect very closely with given choices, the best fitting answer is \( 11 \text{mA} \). Thus, the emitter current is slightly higher than the collector current due to the additional base current flowing in the circuit.

The collector current \( I_c \) in a p-n-p transistor is the main current that gets successfully passed through to the collector. It is often quite close to the emitter current but not exactly the same because a small part of the emitter current siphons off into the base to sustain operations inside the transistor.When we have a situation where 90% of holes emitted reach the collector, the efficiency of conversion from emitter to collector current can be described as:\[I_c = 0.9 \, I_e\]In our p-n-p transistor example, this collector current was given as \( 10 \, \text{mA} \). This is important to note because this current is strictly dependent on how well the transistor can channel the charges from the emitter into the collector. Such efficiency percentages (like 90% in the question) help us understand performance levels of the transistor under specific conditions.By knowing the collector current, and the percentage of the emitter's effectiveness, all calculations concerning related currents become simpler, allowing us to backtrack and confirm other notes, like that of the base or emitter currents, with ease.

The base current \( I_b \) plays an important role in p-n-p transistors but represents a small fraction of the total emitter current. It's the current needed to activate the transistor and keep it operational. In essence, it's like a gate that needs a little bit of current to let a larger current pass through the device. Given the equation for emitter current:\[I_e = I_c + I_b\]We can manipulate this to find the base current:\[ I_b = I_e - I_c\] Based on the calculations from this problem, we previously figured out that the approximate emitter current is \( 11.11 \, \text{mA} \). With the collector current at \( 10 \, \text{mA} \), solving gives:\[I_b = 11.11 \text{mA} - 10 \text{mA} = 1.11 \text{mA}\]This minor current confirms the typical behavior of base currents in p-n-p transistors: small but essential. The slight adjustment to match the given answer options suggests that \( 1 \text{mA} \) is a reasonable estimation of this base current. Understanding this current gives insight into how little energy needs to be diverted from the emitter current to control the much larger flow to the collector.