The equivalent resistance of the circuit shown in the figure is ............. $\Omega$
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(c) The given circuit can be redrawn as follows
$ \Rightarrow $ ${R_{eq}} = 5\,\Omega $.
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A $10\,m$ long potentiometer wire has a potential gradient of $0.0025\, V/cm$. Calculate the distance of null point when the wire is connected to a $1.025\,V$ standard cell ............... $\mathrm{m}$
Two electric bulbs rated ${P_1}\,watt$ $V\, volts$ and ${P_2}\, watt$ $V\, volts$ are connected in parallel and $V\, volts$ are applied to it. The total power will be
In the given figure, battery $E $ is balanced on $55\, cm$ length of potentiometer wire but when a resistance of $10 \,\Omega$ is connected in parallel with the battery then it balances on $50\, cm$ length of the potentiometer wire then internal resistance $r$ of the battery is ............. $\Omega $
In the circuit shown the cells $A$ and $B$ have negligible resistance. For $V _{ A }=12\; V , R _{1}=500\; \Omega$ and $R =100\; \Omega$ the galvanometer $(G)$ shows no deflection. The value of $V_{B}$ is .... $V$
In the circuit shown the resistance of voltmeter is $10,000\, ohm$ and that of ammeter is $20\,ohm$. The ammeter reading is $0.10\,Amp$ and voltmeter reading is $12$ $\mathrm{volt}.$ Then $R$ is equal to .............. $\Omega$
The circuit shown here is used to compare the $e.m.f.$ of two cells ${E_1}$ and ${E_2}({E_1} > {E_2})$. The null point is at $C$ when the galvanometer is connected to ${E_1}$. When the galvanometer is connected to ${E_2}$, the null point will be