The cut-off voltage of the diodes (shown in figure) in forward bias is \(0.6\; V\). The current through the resistor of \(40\) \(Ω\) is____ mA.
Correct Answer: 4
Solution and Explanation
\(D_1 : conducting\) and \(D_2 : open circuit\)
\(⇒i=\frac{1−0.6}{60+40}A\)
=\(\frac{0.4}{100}A\)
\(⇒ i = 4 \;mA\)
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Concepts Used:
Resistance
Resistance is the measure of opposition applied by any object to the flow of electric current. A resistor is an electronic constituent that is used in the circuit with the purpose of offering that specific amount of resistance.
R=V/I
In this case,
v = Voltage across its ends
I = Current flowing through it
All materials resist current flow to some degree. They fall into one of two broad categories:
- Conductors: Materials that offer very little resistance where electrons can move easily. Examples: silver, copper, gold and aluminum.
- Insulators: Materials that present high resistance and restrict the flow of electrons. Examples: Rubber, paper, glass, wood and plastic.
Resistance measurements are normally taken to indicate the condition of a component or a circuit.
- The higher the resistance, the lower the current flow. If abnormally high, one possible cause (among many) could be damaged conductors due to burning or corrosion. All conductors give off some degree of heat, so overheating is an issue often associated with resistance.
- The lower the resistance, the higher the current flow. Possible causes: insulators damaged by moisture or overheating.