In a transistor `10^(8)` electrons enter at the emitter in `10^(-4)` s, out of which `2%` electron go to the base. The current transfer ratio in common base configuration is

In an n-p-n transistor 10^(10) electrons enter the emitter in 10^(-6) s, 3% of the electrons are lost in the base. The current transfer ratio is

In a npn transistor 10^(10) electrons enter the emitter in 10^(–6) s. 4% of the electrons are lost in the base. The current transfer ratio will be

In a n-p-n transistor 10^(10) electrons enter the emitter in 10^(-6)s . 2% of the elecrons are lost in the base. The current transfer ratio and the current amplification factor will be

In an n-p-n transistor 10^(10) electrons enter the emitter in 10^(-6) s. If 2% of the electrons are lost in the base, find the current transfer ratio and the current amplification factor.

In an n-p-n transistor 10^(10) electrons enter the emitter in 10^(-6) s. If 2% of the electrons are lost in the base, find the current transfer ratio and the current amplification factor.

In an n-p-n transistor 10^10 electrons enter the emitter in 10^-6 s . 2% of electrons are lost in the base. Calculate the current transfer ratio and current amplification factor.