Power $P$ is to be delivered to a device via transmission cables having resistance $R_c$. If $V$ is the voltage across $R$ and $I$ the current through it, find the power wasted and how can it be reduced.
The power consumption in transmission lines is given by $P=i^2 R_c$, where $R_c$ is the resistance of transmission lines. The power is given by
$$P=V I$$
The given power can be transmitted in two ways namely (i) at low voltage and high current or (ii) high voltage and low current. In power transmission at low voltage and high current more power is wasted as $P \propto i^2$ whereas power transmission at high voltage and low current facilitates the power transmission with minimal power wastage.
The power wastage can be reduced by transmitting power at high voltage.
$A B$ is a potentiometer wire (figure). If the value of $R$ is increased, in which direction will the balance point J shift?
With the increase of $R$, the current in main circuit decreases which in turn, decreases the potential difference across $A B$ and hence potential gradient $(k)$ across $A B$ decreases.
Since, at neutral point, for given emf of cell, $I$ increases as potential gradient $(k)$ across $A B$ has decreased because
$$E=k I$$
Thus, with the increase of $I$, the balance point neutral point will shift towards $B$.
While doing an experiment with potentiometer (figure) it was found that the deflection is one sided and (i) the deflection decreased while moving from one and $A$ of the wire, to the end $R$; (ii) the deflection increased, while the jockey was moved towards the end $D$.
(i) Which terminal positive or negative of the cell $E_1$ is connected at $X$ in case (i) and how is $E_1$, related to $E$ ?
(ii) Which terminal of the cell $E_1$ is connected at $X$ in case (1 in 1 )?
(i) The deflection in galvanometer is one sided and the deflection decreased, while moving from one end ' $A$ ' of the wire to the end ' $B$ ', thus imply that current in auxiliary circuit (lower circuit containing primary cell) decreases, while potential difference across $A$ and jockey increases.
This is possible only when positive terminal of the cell $E_1$, is connected at $X$ and $E_1>E$.
(ii) The deflection in galvanometer is one sided and the deflection increased, while moving from one end $A$ of the wire to the end $B$, this imply that current in auxiliary circuit (lower circuit containing primary cell) increases, while potential difference across $A$ and jockey increases.
This is possible only when negative terminal of the cell $E_1$, is connected at $X$.
A cell of emf $E$ and internal resistance $r$ is connected across an external resistance $R$. Plot a graph showing the variation of potential differential across $R$, versus $R$.
The graphical relationship between voltage across R and the resistance R is given below
First a set of $n$ equal resistors of $R$ each are connected in series to a battery of emf $E$ and internal resistance $R, A$ current $I$ is observed to flow. Then, the $n$ resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is ' $n$ ' ?
In series combination of resistors, current $I$ is given by $I=\frac{E}{R+n R^{\prime}}$
whereas in parallel combination current $10 I$ is given by $$\frac{E}{R+\frac{R}{n}}=10 I$$
Now, according to problem,
$$\frac{1+n}{1+\frac{1}{n}} \Rightarrow 10=\left(\frac{1+n}{n+1}\right) n \Rightarrow n=10$$