Embibe Experts Solutions for Chapter: Current Electricity, Exercise 3: Exercise-3

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

A ring is made of a wire having a resistance $R_0=12 \mathrm{\Omega }.$ 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 $\frac{8}{3} \mathrm{\Omega }.$

The power dissipated in the circuit shown in the figure is $30 \mathrm{Watts}$. The value of $R$ is 

Cell having an emf $\epsilon$ and internal resistance $r$ is connected across a variable external resistance $R$. As the resistance $R$ is increased, the plot of potential difference $V$ across $R$ is given by, 

A wire of resistance $4 \mathrm{\Omega }$ is stretched to twice its original length. The resistance of stretched wire would be, 

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

The potential difference $\left(V_{\mathrm{A}}-V_{\mathrm{B}}\right)$ between the points $A$ and $B$ in the given figure is

A filament bulb $\left(500 \mathrm{W},100 \mathrm{V}\right)$ is to be used in a $230 \mathrm{V}$ main supply. When a resistance $R$ is connected in series, it works perfectly and the bulb consumes $500 \mathrm{W}.$ The value of $R$ is: