Calculate Total Current \( I_S \):
The total resistance in the circuit is \( 200\Omega + 500\Omega = 700\Omega \). The source voltage is 20V, so the total current \( I_S \) flowing through the series resistance is:
\[ I_S = \frac{20}{700} = \frac{20}{700} \approx 28.6 \text{ mA} \]
Determine Voltage Across the 500 \( \Omega \) Resistor and Zener Diode:
Since the Zener diode is in breakdown mode (10V across it), the voltage drop across the 500 \( \Omega \) resistor is also 10V. The current \( I_1 \) through the 500 \( \Omega \) resistor is:
\[ I_1 = \frac{10}{500} = 20 \text{ mA} \]
Calculate Current Through the Zener Diode \( I_Z \):
The current \( I_Z \) through the Zener diode is:
\[ I_Z = I_S - I_1 = 28.6 - 20 \approx 30 \text{ mA} \]
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).