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$3$ identical bulbs are connected in series and these together dissipate a power $P$. If now the bulbs are connected in parallel, then the power dissipated will be
The galvanometer deflection, when key $K_1$ is closed but $K_2$ is open, equals $\theta_0$ (see figure). On closing $K_2$ also and adjusting $R_2$ to $5\,\Omega $ , the deflection in galvanometer becomes $\frac{{\theta _0}}{5}$. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]: .................. $\Omega$
The resistive network shown below is connected to a $D.C.$ source of $16\, V$. The power consumed by the network is $4\, Watt$. The value of $R$ is ............. $\Omega$
A $9\, V$ battery with internal resistance of $0.5\,\Omega $ is connected across an infinite network as shown in the figure. All ammeters $A_1 , A_2, A_3$ and voltmeter $V$ are ideal. Choose correct statement
Two cells when connected in series are balanced on $8\;m$ on a potentiometer. If the cells are connected with polarities of one of the cell is reversed, they balance on $2\,m$. The ratio of $e.m.f.$'s of the two cells is
A $dc$ source of $emf \,E_1 = 100\,V$ and internal resistance $r = 0.5\,\Omega ,$ a storage battery of emf $E_2 = 90\,V$ and an external resistance $R$ are connected as shown in figure. For what value of $R$ no current will pass through the battery ? ................ $\Omega$
The resistance of an electrical toaster has a temperature dependence given by $R\left( T \right) = {R_0}\left[ {1 + \alpha \left( {T - {T_0}} \right)} \right]$ in its range of operation. At ${T_0} = 300\,K,R = 100\,\Omega $ and at $T = 500\,K,\,R = 120\,\Omega $. The toaster is connected to a voltage source at $200\, V$ and its temperature is raised at a constant rate from $300$ to $500\, K$ in $30\, s$. The total work done in raising the temperature is
Two solid conductors are made up of same material, have same length and same resistance. One of them has a circular cross section of area $A_{1}$ and the other one has a square cross section of area $A_{2}$. The ratio $\frac{A _{1}}{A _{2}}$ is