In the given potentiometer circuit length of the wire $AB$ is $3\, m$ and resistance is $R = 4.5 \, \Omega$ . The length $AC$ for no deflection in galvanometer is
Medium
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since cell connected in secondary circuit is in wrong way so we will not get balancing point.
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The series combination of two batteries, both of the same emf $10 \mathrm{\;V},$ but different internal resistance of $20\; \Omega$ and $5\; \Omega,$ is connected to the parallel combination of two resistors $30\; \Omega$ and $\mathrm{R}\; \Omega .$ The voltage difference across the battery of internal resistance $20\; \Omega$ is zero, the value of $\mathrm{R}(\text { in } \Omega)$ is
In the figure, the value of resistors to be connected between $C$ and $D$ so that the resistance of the entire circuit between $A$ and $B$ does not change with the number of elementary sets used is
Assertion $(A):$ In a meter bridge experiment, null point for an unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same $p$ as before by decreasing the value of the standard resistance.
Reason $(R):$ Resistance of metal increases with increase in temperature.
Power dissipated across the $8 \,\,\Omega$ resistor in the circuit shown here is $2\,\, watt.$ The power dissipated in watt units across the $3 \,\,\Omega$ resistor is
Three resistors having resistances $\mathrm{r}_{1}, \mathrm{r}_{2}$ and $\mathrm{r}_{3}$ are connected as shown in the given circuit. The ratio $\frac{i_{3}}{i_{1}}$ of currents in terms of resistances used in the circuit is :
Voltmeter reads potential difference across the terminals of an old battery as $1.2\,volt$ , while a potentiometer reads $1.4\,volt$ . The internal resistance of battery is $40\,\Omega $ , then voltmeter resistance is .............. $\Omega$