The ratio of the equivalent resistance of the network (shown in figure) between the points $a$ and ${b}$ when switch is open and switch is closed is ${x}: 8 .$ The value of ${x}$ is ..... .
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Two resistances of $400$ $\Omega$ and $800$ $\Omega$ are connected in series with $6\, volt$ battery of negligible internal resistance. A voltmeter of resistance $10,000$ $\Omega$ is used to measure the potential difference across $400$ $\Omega$. The error in the measurement of potential difference in volts approximately is
In the given potentiometer circuit, the resistance of the potentiometer wire $AB$ is $R_0$. $C$ is a cell of internal resistance $r$. The galvanometer $G$ does not give zero deflection for any position of the jockey $J$. Which of the following cannot be a reason for this?
Wheatstone bridge principle is used to measure the specific resistance $\left(S_1\right)$ of given wire, having length $L$, radius $r$. If $X$ is the resistance of wire, then specific resistance is: $S_1=X\left(\frac{\pi r^2}{L}\right)$. If the length of the wire gets doubled then the value of specific resistance will be :
A platinum resistance thermometer has a resistance of $50\,\Omega $ at $20\,^o C$. When dipped in a liquid the resistance becomes $76.8\,\Omega $. The temperature coefficient of resistance for platinum is $\alpha = 3.92 \times {10^{ - 3}}\,^o C$. The temperature of the liquid is .............. $^o C$
In the circuit shown, the current through the $4\,\Omega $ resistor is $1\, amp$ when the points $P$ and $M$ are connected to a d.c. voltage source. The potential difference between the points $M$ and $N$ is ............... $volt$
Two cells of $e.m.f.$ $E_1$ and $E_2$ are joined in series and the balancing length of the potentiometer wire is $625$ $cm$. If the terminals of $E_1$ are reversed, the balancing length obtained is $125 \,cm$. Given $E_2 > E_1$, the ratio $E_1: E_2$ will be
There are a large number of cells available, each marked $(6 \,V , 0.5 \,\Omega)$ to be used to supply current to a device of resistance $0.75 \,\Omega$, requiring $24 \,A$ current. How should the cells be arranged, so that power is transmitted to the load using minimum number of cells?
A potentiometer $PQ$ is set up to compare two resistances as shown in the figure. The ameter $A$ in the circuit reads $1.0\, A$ when two way key $K_3$ is open. The balance point is at a length $l_1\, cm$ from $P$ when two way key $K_3$ is plugged in between $2$ and $1$ , while the balance point is at a length $l_2\, cm$ from $P$ when key $K_3$ is plugged in between $3$ and $1$ . The ratio of two resistances $\frac{{{R_1}}}{{{R_2}}}$ is found to be