
The current does not rise immediately in a circuit containing inductance.
A. because of induced emf
B. because of high voltage drop
C. both 1 and 2
D. because of joule heating
Answer
560.1k+ views
Hint: Inductor is a device which responds to the change in the current. It stores the magnetic energy. When an inductor is connected to the power source initially it won’t allow the flow of current but after some time it will act as a normal conducting wire with no resistance which will be exactly opposite in case of capacitor.
Formula used:
$e = - L\dfrac{{di}}{{dt}}$
Complete answer:
Normally in case of capacitors initially when power source is connected to the capacitor it acts as
a normal conduction wire and after sometime this will completely not allow the flow of charge but in case of an inductor this will be completely opposite. When an inductor is connected to the power source initially it won’t allow the flow of current but gradually after some time it will allow the flow of current just as the normal conducting wire.
The induced emf in an inductor will be $e = - L\dfrac{{di}}{{dt}}$
Initially when the inductor is connected to the voltage source the current tries to rise faster. Since the induced emf in an inductor is opposite to the rate of change of current, if the rate is higher the inductor tries to block the rate by inducing the opposite emf.
So initially the rate of change of current will be higher, so induced emf will be higher and hence it opposes the current flow.
So gradually the rate of increase decreases and voltage source will neutralize the induced emf effect and current flow freely.
Hence the induced emf is the reason for the slower rise of current.
Hence option A will be the answer.
Note:
In case of an inductor when the key is closed then there is sudden increase in the flow of current which means that sudden change will be opposed by the inductor. Hence inductor generates emf itself in such a way that it opposes that flow and hence net current flown will be zero and initially inductor doesn’t allow the current.
Formula used:
$e = - L\dfrac{{di}}{{dt}}$
Complete answer:
Normally in case of capacitors initially when power source is connected to the capacitor it acts as
a normal conduction wire and after sometime this will completely not allow the flow of charge but in case of an inductor this will be completely opposite. When an inductor is connected to the power source initially it won’t allow the flow of current but gradually after some time it will allow the flow of current just as the normal conducting wire.
The induced emf in an inductor will be $e = - L\dfrac{{di}}{{dt}}$
Initially when the inductor is connected to the voltage source the current tries to rise faster. Since the induced emf in an inductor is opposite to the rate of change of current, if the rate is higher the inductor tries to block the rate by inducing the opposite emf.
So initially the rate of change of current will be higher, so induced emf will be higher and hence it opposes the current flow.
So gradually the rate of increase decreases and voltage source will neutralize the induced emf effect and current flow freely.
Hence the induced emf is the reason for the slower rise of current.
Hence option A will be the answer.
Note:
In case of an inductor when the key is closed then there is sudden increase in the flow of current which means that sudden change will be opposed by the inductor. Hence inductor generates emf itself in such a way that it opposes that flow and hence net current flown will be zero and initially inductor doesn’t allow the current.
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