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5/7 A

5/11 A

1 A

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In the above diagram when the key K is closed, we see that a potential difference of 10V and a current has been induced in the circuit, now when the current tries to pass through the 2mH inductor, it could not pass because of its excessively high resistance, so the current takes the path 1, shown in the diagram.

Now in path 1,

Equivalent resistance,

${{R}_{eq}}=\left( 6+8 \right)\Omega $ ,

${{R}_{eq}}=14\Omega $,

So the value of current ${{I}_{1}}$ flowing through the circuit is,

${{I}_{1}}=\dfrac{V}{{{R}_{eq}}}$,

${{I}_{1}}=\dfrac{10}{14}$,

${{I}_{1}}=\dfrac{5}{7}A$,

Resistance is a device that is used in the electrical circuit to stop the flow of current inside the circuit sometimes partially and sometimes fully.

The difference between the current at two points of an electrical circuit is known as potential difference.

We can show the potential difference by a simple law known as ohm’s law, which states that the voltage or potential difference of a circuit is proportional to the amount of current flowing through the circuit and the resistance offered.

Inductor is an electrical component that has a very high resistance and its work is to oppose the flow of current through the circuit. Hence the current is passing through a loop named path 1.

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