The potential difference between point $A$ and $B$ is ............. $V$
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(d) The given circuit is a balanced wheatstone bridge circuit. Hence potential difference between $A$ and $B$ is zero.
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Resistances of $6\, ohm$ each are connected in the manner shown in adjoining figure. With the current $0.5\,ampere$ as shown in figure, the potential difference ${V_P} - {V_Q}$ is .............. $V$
The number density of free electrons in copper is nearly $8 \times 10^{28}\,m ^{-3} . A$ copper wire has its area of cross section $=2 \times 10^{-6}\,m ^2$ and is carrying a current of $3.2\,A$. The drift speed of the electrons is $.....\times 10^{-6}\,ms ^{-1}$.
Consider a wire having current $10\,A$ having area of crossection $1\,cm^2$. If number of electrons per unit volume is $9 \times 10^{28}\, m^{-3}$. Find the drift velocity of electrons
In the given figure $R_1=10 \Omega, R_2=8 \Omega, R_3=4 \Omega$ and $R_4=8 \Omega$. Battery is ideal with emf $12 \mathrm{~V}$. Equivalent resistant of the circuit and current supplied by battery are respectively.
In the circuit shown, the reading of the Ammeter is doubled after the switch is closed. Each resistor has a resistance $ = 1\,\Omega $ and the ideal cell has an $e.m.f. = 10\, V$. Then, the Ammeter has a coil resistance equal to ................ $\Omega$
Consider the circuits shown in the figure. Both the circuits are taking same current from battery but current through $R$ in the second circuit is $\frac{1}{{10}}$$^{th}$ of current through $R$ in the first circuit. If $R$ is $11$ $\Omega$, the value of $ R_1$ ................ $\Omega$