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For a cell of $e.m.f.$ $2\,V$, a balance is obtained for $50\, cm$ of the potentiometer wire. If the cell is shunted by a $2\,\Omega $ resistor and the balance is obtained across $40\, cm$ of the wire, then the internal resistance of the cell is ............. $\Omega $
In order to measure the internal resistance $r_1$ of a cell of emf $E$, a meter bridge of wire resistance $R_0=50 \Omega$, a resistance $R_0 / 2$, another cell of emf $E / 2$ (internal resistance $r$ ) and a galvanometer $G$ are used in a circuit, as shown in the figure. If the null point is found at $l=72 cm$, then the value of $r_1=$ . . . . $\Omega$
$50\,\Omega $ and $100\,\Omega $ resistors are connected in series. This connection is connected with a battery of $2.4\, volts$. When a voltmeter of $100\,\Omega $ resistance is connected across $100\,\Omega $ resistor, then the reading of the voltmeter will be ............. $V$
In this figure the resistance of the coil of galvanometer $G$ is $2\,\Omega$. The emf of the cell is $4\,V$. The ratio of potential difference across $C_1$ and $C_2$ is:
The circuit diagram shown consists of a large number of element (each element has two resistors $R_1$ and $R_2$). The resistance of the resistors in each subsequent element differs by $a$ factor of $K = 1/2$ from the resistance of the resistors in the previous elements. The equivalent reistance between $A$ and $B$ shown in figure is :