A current of two amperes is flowing through a cell of $e.m.f.$ $5\, volts$ and internal resistance $0.5\, ohm$ from negative to positive electrode. If the potential of negative electrode is $10\,V$, the potential of positive electrode will be .............. $V$
A$5$
B$14 $
C$15$
D$16$
Medium
Download our app for free and get started
B$14 $
b ${V_2} - {V_1} = E - ir = 5 - 2 \times 0.5 = 4\,volt$
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A source of $e.m.f.$ $E = 15\,V$ and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time as $i = 1.2 t + 3$. Then, the total charge that will flow in first five second will be ............... $C$
In the given figure, the $emf$ of the cell is $2.2\, {V}$ and if internal resistance is $0.6\, \Omega$. Calculate the power dissipated in the whole circuit: (in $W$)
The resistance of a wire is $R$. It is bent at the middle by $180^{\circ}$ and both the ends are twisted together to make a shorter wire. The resistance of the new wire is
An electric wire of length ‘$I$’ and area of cross-section $a$ has a resistance $R\, ohms$. Another wire of the same material having same length and area of cross-section $4a$ has a resistance of
If a wire of resistance $20\,\Omega $ is covered with ice and a voltage of $210\, V$ is applied across the wire, then the rate of melting of ice is .................. $g/s$
In the circuit shown in figure reading of voltmeter is $V_1$ when only $S_1$ is closed, reading of voltmeter is $ V_2$ when only $S_2$ is closed and reading of voltmeter is $V_3$ when both $S_1$ and $S_2$ are closed. Then