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An ammeter with internal resistance $90\,\Omega $ reads $1.85\, A$ when connected in a circuit containing a battery and two resistors $700\,\Omega $ and $410\,\Omega $ in series. Actual current will be
An unknown resistance $R_1$ is connected in series with a resistance of $10 \,\Omega$. This combinations is connected to one gap of a meter bridge while a resistance $R_2$ is connected in the other gap. The balance point is at $50\, cm$. Now, when the $10 \,\Omega$ resistance is removed the balance point shifts to $40\, cm$. The value of $R_1$ is (in $ohm$)
In an experiment of potentiometer for measuring the internal resistance of primary cell a balancing length $\ell $ is obtained on the potentiometer wire when the cell is open circuit. Now the cell is short circuited by a resistance $R$. If $R$ is to be equal to the internal resistance of the cell the balancing length on the potentiometer wire will be
A meter bridge set up as shown to determine end correction at $A$ and $B$ . When a resistance of $15\,\Omega $ is used in left gap and of $20\,\Omega $ in right gap, then null point comes at a distance $42\ cm$ from $A$ . When these resistances are interchanged null point comes at a distance $57\ cm$ from $A$ . Values of end corrections are
In the figure shown for gives values of $R_1$ and $R_2$ the balance point for Jockey is at $40\,cm$ from $A$. When $R_2$ is shunted by a resistance of $10\, \Omega$ , balance shifts to $50\,cm.$ $R_1$ and $R_2$ are $(AB = 1 \,m)$
In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point $B$ with respect to the point $A$ is$.......V$