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Wheatstone bridge principle is used to measure the specific resistance $\left(S_1\right)$ of given wire, having length $L$, radius $r$. If $X$ is the resistance of wire, then specific resistance is: $S_1=X\left(\frac{\pi r^2}{L}\right)$. If the length of the wire gets doubled then the value of specific resistance will be :
If you are provided a set of resistances $2\, \Omega, 4\, \Omega$ $6\, \Omega$ and $8\, \Omega$. Connect these resistances so as to obtain an equivalent resistance of $\frac{46}{3}\, \Omega$.
The resistance ofeach arm of the Wheatstone's bridge is $10\; ohm$. A resistance of $10 \;ohm$ is connected in series with a galvanometer then the equivalent resistance across the battery will be
The circuit shown in the figure consists of a battery of $emf$ $\varepsilon = 10 \,V$ ; a capacitor of capacitance $C = 1.0$ $ \mu F$ and three resistor of values $R_1 = 2$ $\Omega$ , $R_2 = 2$ $\Omega$ and $R_3 = 1$ $\Omega$ . Initially the capacitor is completely uncharged and the switch $S$ is open. The switch $S$ is closed at $t = 0.$
If an electron revolves in the path of a circle of radius of $0.5 × 10^{-10}\, m$ at frequency of $5 × 10^{15}$ $cycles/s$ the electric current in the circle is ..................$mA$ (Charge of an electron $=1.6 × 10^{-19}\, C$ )