In order to measure the internal resistance $r_1$ of a cell of emf $E$, a meter bridge of wire resistance $R_0=50 \Omega$, a resistance $R_0 / 2$, another cell of emf $E / 2$ (internal resistance $r$ ) and a galvanometer $G$ are used in a circuit, as shown in the figure. If the null point is found at $l=72 cm$, then the value of $r_1=$ . . . . $\Omega$
$i=\frac{E_0}{2 \times 0.78 R_0}=\frac{E_0}{ r _1+\frac{3}{2} R _0}$
$r _1+1.5 R _0=1.56 R _0$
$r _1=0.06 R _0$
$=0.06 \times 50=3 \Omega$
Download our app
and get started for free
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.*
Drift speed of electrons, when $1.5\, A$ of current flows in a copper wire of cross section $5\, mm^2$, is $v$. If the electron density in copper is $9 \times 10^{28}\, m^3$ the value of $v$ in $mm/s$ is close to (Take charge of electron to be $= 1.6 \times 10^{-19}\, C$)
A $10 \,m$ long potentiometer wire is connected to a battery having a steady voltage. A Leclanche cell is balanced at $4 \,m$ length of the wire. If the length is kept the same, but its cross-section is doubled, the null point will be obtained at ........... $m$
In an electric circuit, a cell of certain emf provides a potential difference of $1.25\, {V}$ across a load resistance of $5\, \Omega .$ However, it provides a potential difference of $1\, {V}$ across a load resistance of $2\, \Omega$. The $emf$ of the cell is given by $\frac{x}{10} v$. Then the value of $x$ is ..... .
In given arrangement $E_1 = 5\, volts$ $E_2 = 7\, volt$ balancing length is $6\,m$ if terminals of $E_2$ are reversed then new balancing length will be
The charge of an electron is $1.6 × 10^{-19}\,C$. How many electrons strike the screen of a cathode ray tube each second when the beam current is $16\,mA$
A battery has $e.m.f.$ $4\, V$ and internal resistance $r$. When this battery is connected to an external resistance of $2\, ohms$, a current of $1\, amp$. flows in the circuit. How much current will flow if the terminals of the battery are connected directly .......... $amp$