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ICSE Class 10 Physics Question 2 of 31

Current Electricity — Question 1

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Question 1

In a conductor, 6.25 × 1016 electrons flow from it's end A to B in 2 s. Find the current flowing through the conductor. (e = 1.6 × 10-19 C)

Answer

Given,

Number of electrons flowing (n) = 6.25 × 1016

Time taken (t) = 2 s

e = 1.6 × 10-19 C

Current in conductor is given by,

I = net\dfrac{\text{ne}}{\text{t}}

Substituting the values in the formula above we get,

I=6.25×1016×1.6×10−192I=6.25×1.6×10−32I=6.25×0.8×10−3I=5×10−3AI = \dfrac{6.25 \times 10^{16} \times 1.6 \times 10^{-19}}{2} \\[0.5em] I = \dfrac{6.25 \times 1.6 \times 10^{-3}}{2} \\[0.5em] I = 6.25 \times 0.8 \times 10^{-3} \\[0.5em] I = 5 \times 10^{-3} A

Hence, the current flowing through the conductor = 5 mA from B to A