
For zero galvanometer deflection, the bridge must be balanced. This is a Wheatstone bridge arrangement.
The condition for a balanced Wheatstone bridge is:
$\frac{R_1}{R_2} = \frac{R_3}{R_4}$
Where R1, R2, R3, and R4 represent the resistance in each arm of the Wheatstone bridge. Here, R1 and R2 are the resistances of the wire segments AJ and JB respectively, and R3 = 8 Ω and R4 = 12 Ω.
The length of AB is 40 cm, thus, $\frac{R_1}{R_2} = \frac{8}{12} = \frac{2}{3}$
$\frac{R_1}{R_2} = \frac{x}{40 - x} = \frac{2}{3}$
3x = 80 − 2x 5x = 80 x = 16 cm
Distance of J from B = 40 cm − 16 cm = 24 cm
An air filled parallel plate electrostatic actuator is shown in the figure. The area of each capacitor plate is $100 \mu m \times 100 \mu m$. The distance between the plates $d_0 = 1 \mu m$ when both the capacitor charge and spring restoring force are zero as shown in Figure (a). A linear spring of constant $k = 0.01 N/m$ is connected to the movable plate. When charge is supplied to the capacitor using a current source, the top plate moves as shown in Figure (b). The magnitude of minimum charge (Q) required to momentarily close the gap between the plates is ________ $\times 10^{-14} C$ (rounded off to two decimal places). Note: Assume a full range of motion is possible for the top plate and there is no fringe capacitance. The permittivity of free space is $\epsilon_0 = 8.85 \times 10^{-12} F/m$ and relative permittivity of air ($\epsilon_r$) is 1.
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The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.