A bulb of $100\, W$ is connected in parallel with an ideal inductance of $1\, H$. This arrangement is connected to a $90\,V$ battery through a switch. On pressing the switch, the
A
bulb does not glow
B
bulb glows
C
bulb glows after a short time and then continues to glow
D
bulb glows for a short time and then stops glowing.
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A
bulb does not glow
a inductance will act like a short circuit, so whole of the current will pass through the inductance and no current will flow through the bulb. Therefore, the bulb will not glow.
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A $5\, V$ battery with internal resistance $2\, \Omega$ and a $2\,V$ battery internal resistance $1\, \Omega$ are connected to a $10\, \Omega$ resistor as shown in the figure. The current in the $10\, \Omega$ resistor is :-
The adjoining figure shows the connections of potentiometer experiment to determine internal resistance of of a leclanche cell. When the cell is on open circuit the balancing length of the potentiometer wire is $3.4\, m$ and on closing the key $K_2$ the balancing length becomes $1.7\, m$ . If the resistance $R$ through which current is drawn is $10\,\Omega $ then the internal resistance of the cell is .............. $\Omega$