Current Electricity Questions

For ensuring dissipation of same energy in all three resistors ($R_1,R_2,R_3$) connected as shown in figure, their values must be related as:

In producing chlorine through electrolysis $100\text{W}$ power at $125\text{V}$ is being consumed. How much chlorine per min is liberated? ECE of chlorine is $0.367\times 10^{-6}\text{kg/C}$:

Consider an electric circuit consisting of two batteries with emfs ${ϵ}_1,{ϵ}_2$ and internal resistances $r_1,r_2$ connected in parallel with an external resistance $R$. If $i_1$ and $i_2$ are currents through the respective branches, which of the following Kirchhoff equations is correct?

A galvanometer of resistance $50\Omega$ is connected to a battery of $3\text{V}$ along with a resistance of $2950\Omega$ in series. A full scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $20$ divisions, the resistance in series should be:

A current of $3\text{A}$ flows through the $2\Omega$ resistor shown in the circuit. The power dissipated in the $5\Omega$ resistor is:

The power dissipated across the $8\Omega$ resistor in the circuit shown here is $2\text{W}$. The power dissipated in watt units across the $3\Omega$ resistor is :(Assume a circuit where the $8\Omega$ and $24\Omega$ resistors are in parallel, and their combination is in series with a $3\Omega$ resistor)

In a circuit, the current through a $4\Omega$ resistor is $1\text{A}$ when the points $P$ and $M$ are connected to a d.c. voltage source. The potential difference between the points $M$ and $N$ is:

A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively:

A cell can be balanced against $110\text{cm}$ and $100\text{cm}$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10\Omega$. Its internal resistance is:

A thermocouple of negligible resistance produces an $e.m.f.$ of $40\mu {\text{V/}}^{\circ }\text{C}$ in the linear range. A galvanometer of resistance $10\Omega$ with sensitivity $1\mu \text{A/div}$ is used. The smallest temperature difference detectable is:

In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will: (All resistance given in ohms)

The potential difference $(V_A-V_B)$ between points $A$ and $B$ in the given figure is:

Two cells having the same emf, are connected in series through an external resistance $R$. Cells have internal resistance $r_1$ and $r_2$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is:

Kirchhoff’s first and second laws for electrical circuits are consequences of:

The resistances of the four arms $P,Q,R,S$ in a Wheatstone bridge are $10\Omega ,30\Omega ,30\Omega ,90\Omega$ respectively. The e.m.f. is $7\text{V}$ and internal resistance is $5\Omega$. What is the current drawn from the cell?

A small signal voltage $V(t)=V_0\sin \omega t$ is applied across an ideal capacitor $C$:

The charge flowing through a resistance $R$ varies with time $t$ as $Q=at-b{t}^2$ where $a$ and $b$ are positive constants. The total heat produced in $R$ is:

A potentiometer wire is $100\text{cm}$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50\text{cm}$ and $10\text{cm}$ from the positive end of the wire in the two cases. The ratio of emf’s is:

A circuit contains an ammeter, a battery of $30\text{V}$ and a resistance $40.8\Omega$ all connected in series. If the ammeter has a coil of resistance $480\Omega$ and a shunt of $20\Omega$, the reading in the ammeter will be:

A series R-C circuit is connected to an alternating voltage source. Consider two situations:(i) When capacitor is air filled.(ii) When capacitor is mica filled.Current through resistor is $i$ and voltage across capacitor is $V$ then:

Two metal wires of identical dimensions are connected in series. If ${\sigma }_1$ and ${\sigma }_2$ are the conductivities of the metal wires respectively, the effective conductivity of the combination is:

A potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. $E$ is balanced at a length $l$ of the potentiometer wire. The e.m.f. $E$ will be given by:

In an ammeter $0.2\%$ of main current passes through the galvanometer. If resistance of galvanometer is $G$, the resistance of ammeter will be:

A potentiometer circuit has been set up for finding the internal resistance of a given cell. When the resistance, $R$, connected across the given cell, has values of (i) infinity and (ii) $9.5\Omega$, the 'balancing lengths' are found to be $3\text{m}$ and $2.85\text{m}$, respectively. The value of internal resistance of the cell is:

The resistance in the two arms of the meter bridge are $5\Omega$ and $R\Omega$, respectively. When the resistance $R$ is shunted with an equal resistance, the new balance point is at $1.6l_1$. The resistance '$R$ ', is:

Two cities are $150\text{km}$ apart. Electric power is sent from one city to another city through copper wires. The fall of potential per $\text{km}$ is $8\text{volt}$ and the average resistance per $\text{km}$ is $0.5\Omega$. The power loss in the wire is:

The mean free path of electrons in a metal is $4\times 10^{-8}\text{m}$. The electric field which can give on an average $2\text{eV}$ energy to an electron in the metal will be in units of $\text{V/m}$:

The internal resistance of a $2.1\text{V}$ cell which gives a current of $0.2\text{A}$ through a resistance of $10\Omega$ is:

A wire of resistance $4\Omega$ is stretched to twice its original length. The resistance of the stretched wire would be:

A milli voltmeter of $25\text{mV}$ range is to be converted into an ammeter of $25\text{A}$ range. The value (in ohm) of necessary shunt will be:

If voltage across a bulb rated $220\text{Volt}-100\text{Watt}$ drops by $2.5\%$ of its rated value, the percentage of the rated value by which the power would decrease is:

In the circuit shown the cells $A$ and $B$ have negligible resistances. For $V_A=12\text{V},R_1=500\Omega$ and $R=100\Omega$ the galvanometer ($G$) shows no deflection. The value of $V_B$ is:

A ring is made of a wire having a resistance $R_0=12\Omega$. Find the points $A$ and $B$, as shown in the figure, at which a current carrying conductor should be connected so that the resistance $R$ of the sub circuit between these points is equal to $8/3\Omega$.

A cell with emf $ϵ$ and internal resistance $r$ is connected to variable $R$. Plot of potential difference $V$ across $R$ vs $R$ is:

The power dissipated in the circuit shown (two $10\Omega$ resistors in parallel connected to $R$ in series, $10\text{V}$ battery) is $30\text{Watts}$. The value of $R$ is:

In the circuit shown, if $V_A=0$, what is the potential at point $B$?

An inductor $20\text{mH}$, a capacitor $50\mu \text{F}$ and a resistor $40\Omega$ are connected in series across a source of emf $V=10\sin 340t$. The power loss in A.C. circuit is:

A galvanometer of resistance $G$ is shunted by a resistance $S$ ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is: