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
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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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Three resistors are connected to form the sides of a triangle $ABC$, the resistance of the sides $AB$, $BC$ and $CA$ are $40\,ohms$, $60\,ohms$ and $100\,ohms$ respectively. The effective resistance between the points $A$ and $B$ in $ohms$ will be
In the circuit shown in figure, potential difference between points $A$ and $B$ is $16\, V$. The current passing through $2\,\Omega $ resistance will be ................. $\mathrm{A}$
Two wires of the same dimensions but resistivities ${\rho _1}$ and ${\rho _2}$ are connected in series. The equivalent resistivity of the combination is
If power dissipated in the $9 \,\Omega$ resistor in the circuit shown is $36\,W$, the potential difference across the $2 \,\Omega$ resistor is .......... $V$
In the given circuit the internal resistance of the $18\,V$ cell is negligible. If $R_1 = 400 \,\Omega ,\,R_3 = 100\,\Omega $ and $R_4 = 500\,\Omega $ and the reading of an ideal voltmeter across $R_4$ is $5\,V,$ then the value of $R_2$ will be ........... $\Omega$