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Find the Current Through the Battery in Each of the Circuits Shown in Figure.

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प्रश्न

Find the current through the battery in each of the circuits shown in figure.

(Assume that the resistance of each diode is zero in forward bias and is infinity in reverse bias.)

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उत्तर

We know that when a diode is forward biassed, it has zero resistance ideally. So, it can be replaced by a short circuit. When a diode is reverse biassed, it has infinite resistance ideally. So, it can be replaced by an open circuit.


(a) In the given circuit diagram, both diodes are forward biassed. So, the resistance of both of them is zero. Thus, the diode resistance is zero.

\[i = \frac{5}{\frac{10 \times 10}{10 + 10}} = \frac{5}{5} = 1  A\]

(b) One diode is forward biassed and the other is reverse biassed. The reverse-biassed diode is replaced by an open circuit, so no current flows through this branch.
The current passes through the forward-biassed diode only.

\[i = \frac{V}{R_{net}}   =   \frac{5}{10} = 0 . 5  A\]

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अध्याय 45: Semiconductors and Semiconductor Devices - Exercises [पृष्ठ ४२०]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 45 Semiconductors and Semiconductor Devices
Exercises | Q 27 | पृष्ठ ४२०

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In an unbiased p-n junction, holes diffuse from the p-region to n-region because ______.


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I = `"I"_0 exp ("eV"/(2"k"_"BT") - 1)`

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(a) What will be the forward current at a forward voltage of 0.6 V?

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(c) What is the dynamic resistance?

(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?


Write the two processes that take place in the formation of a p-n junction.


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Explain, with the help of a circuit diagram, the working of n-p-n transistor as a common emitter amplifier.


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