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JEE Main 202124 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Morning Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2021 · 24 February · Morning Shift · Physics · Semiconductors

In connection with the circuit drawn below, the value of current...

In connection with the circuit drawn below, the value of current flowing through 2kΩ2kΩ resistor is______×10−4A.\times {10}^{- 4}A .

In connection with the circuit drawn below, the value of current flowing through 2kΩ resistor is______ × 10 ^ - 4 A .

Correct answer

0

Step-by-step solution

Current through 2kΩ2kΩ resistance

I=52×103=2.5×10−3AI = \dfrac{5}{2 \times {10}^{3}} = 2 . 5 \times {10}^{- 3}A

I=25×10−4AI = 25 \times {10}^{- 4}A

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