In the circuit shown in the figure, no current flows through the ideal ammeter. If the internal resistance of the cell is negligible, the value of unknown resistance $R $ is .............. $\Omega$
Medium
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If no current flows through the ammeter $(1),$ potential on the $20 \mathrm{\,W}$ resistance becomes $2 \mathrm{\,V}.$
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The resistance of a wire is ${10^{ - 6}}\,\Omega $ per meter. It is bend in the form of a circle of diameter $2\,m$. A wire of the same material is connected across its diameter. The total resistance across its diameter $AB$ will be
For a cell terminal $P.D.$ is $2.2\;V$ when circuit is open and reduces to $1.8\;V$ when cell is connected to a resistance of $R = 5\,\Omega $. Determine internal resistance of cell $(r)$ is then ........ $\Omega$
Refer to the circuit shown. What will be the total power dissipation in the circuit if $P$ is the power dissipated in $R_1$ ? It is given that $R_2=4 R_1$ and $R_3=12 R_1$ are .......... $P$
Assertion : Free electrons always keep on moving in a conductor even then no magnetic force act on them in magnetic field unless a current is passed through it.
Reason : The average velocity of free electron is zero.
The meter bridge shown is in balanced position with $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}}$. If we now litterchange the positions of gavanometer and cell, will the bridge work? If yes, what will be balance condition?