Discussion
The current 'I' supplied by the power supply in the given circuit is ...............
Answer: C
The current is 0.1 A.
Conventional current leaves the positive (left) plate of the cell, passes the 20 Ω resistor and reaches the left junction. There it meets two diodes. The middle diode points from left to right, so it is forward biased and, taken as ideal, acts as a short. The top diode points from right to left, so it is reverse biased and blocks its branch, 50 Ω resistor and all.
The current therefore flows through the middle diode and the 30 Ω resistor back to the cell:
\[ \begin{aligned} I &= \frac{5\ \text{V}}{20\ \Omega + 30\ \Omega} \\ &= 0.1\ \text{A} \end{aligned} \]
0.05 A is what one gets by sending the current through the 50 Ω branch as well (5/100), which the reversed top diode does not allow. 1 mA and 5 mA are far too small for a 5 V source driving 50 Ω.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development