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A battery of $24$ cells, each of emf $1.5\, V$ and internal resistance $2\, \Omega$ is to be connected in order to send the maximum current through a $12 \,\Omega$ resistor. The correct arrangement of cells will be
A resistance of $4\,\Omega $ and a wire of length $5\,m$ and resistance $5\,\Omega $ are joined in series and connected to a cell of $e.m.f.$ $10\, V$ and internal resistance $1\,\Omega $. A parallel combination of two identical cells is balanced across $300\, cm$ of the wire. The $e.m.f.$ $E$ of each cell is ........... $V$
Current $l$ versus time $t$ graph through a conductor is shown in the figure. Average current through the conductor in the interval $0$ to $15 \,s$ is ............ $A$
In a potentiometer arrangement, a cell of emf $1.20\, V$ gives a balance point at $36\, cm$ length of wire. This cell is now replaced by another cell of emf $1.80\, V$. The difference in balancing length of potentiometer wire in above conditions will be $....cm$.
Four ammeters with identical internal resistances $r$ and a resistor of resistance $R$ are connected to a current source as shown in figure. It is known that the reading of the ammeter $A_1$ is $I_1 = 3\ A$ and the reading of the ammeter $A_2$ is $I_2 = 5\ A$ . Determine the ratio of the resistances $R/r$ .
A potential $V_0$ is applied across a uniform wire of resistance $R$. The power dissipation is $P_1$. The wire is then cut into two equal halves and a potential of $V _0$ is applied across the length of each half. The total power dissipation across two wires is $P_2$. The ratio $P_2: P_1$ is $\sqrt{x}: 1$. The value of $x$ is $.............$.