Question
Question: The current through a wire depends on time as \(i={{i}_{0}}+\alpha t\), where \({{i}_{0}}=10A\) and ...
The current through a wire depends on time as i=i0+αt, where i0=10A and α=4A/s. Find the charge crossed through a section of the wire in 10 seconds.
Solution
Write current as differential rate of flow of charge. Integrate the equation to obtain an expression for the total charge crossed through the section of wire in that time period. Substitute the given values to obtain the magnitude of charge crossed.
Complete answer:
Electric current is defined as rate of transference of charge. We can write electric current as
I=dtdQ where Q represents charge and t represents time.
This implies that,
dQ=Idt
The electric current through the wire is given as a function of time as i=i0+αt. Substituting this value in above equation, we have
dQ=(i0+αt)dt
To find the charge crossed through a section of the wire in 10 seconds, we integrate this equation from t = 0 to t = 10 sec. By doing so we get, charge crossed in given time interval
Q=∫dQ=0∫10Idt
Q=0∫10i0dt+αtdt=[i0t+2αt2]010
Q=10i0+2α(10)2=10i0+50α
It is given that i0=10A and α=4A/s. Substituting these values, we have
Q=10×10+50×4=300As
Since, 1As=1C Therefore,
Q=300C
The charge crossed through a section of the wire in 10 seconds is 300 coulombs.
Additional Information:
The relation of current and voltage is given by ohm’s law which states that current through a conductor is linearly proportional to the voltage applied. Mathematically,
V∝I⇒V=IR
The proportionality constant R is known as resistance of the conductor.
Electric current can be measured by an instrument known as ammeter.
Moving charges or electric current produces magnetic fields. It can be said that moving charges produce both electric and magnetic fields.
Note:
Electric current is defined as rate of transference of charge and can be written as I=dtdQ
One ampere of current is the rate of flow of one coulomb of charge in one second.
Students must learn basic formulas of integration and differentiation as these are very useful in solving many problems.