The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be _____ A.
Since the potential at \( A \) and \( B \) is the same, no current flows through the middle resistor (30 \( \Omega \)).
The two parallel branches consist of: \[ R_1 = 10 \Omega + 20 \Omega = 30 \Omega \] \[ R_2 = 40 \Omega \] The equivalent resistance: \[ R_{{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(30)(40)}{30 + 40} = \frac{1200}{70} = 17.14 \Omega \] The total current using Ohm’s law: \[ I = \frac{40V}{17.14} = 2.33 A \]
In the circuit shown, assuming the threshold voltage of the diode is negligibly small, then the voltage \( V_{AB} \) is correctly represented by:
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.