Step 1: Finding Resistance (\(R\))
We use the formula for resistance when the rated voltage (\(V\)) and rated power (\(P\)) are given:
\(R = \frac{V^2}{P}\)
Given:
\(V = 200 \, \text{volts}, \quad P = 50 \, \text{watts}\)
Substituting the values:
\(R = \frac{200^2}{50} = \frac{40000}{50} = 800 \, \Omega\)
Thus, the resistance is:
\(R = 800 \, \Omega\)
Step 2: Finding Power (\(P\)) for a Different Voltage
To calculate the power consumed for an applied voltage (\(V_{\text{applied}}\)) of 100 volts, we use the formula:
\(P = \frac{V_{\text{applied}}^2}{R}\)
Given:
\(V_{\text{applied}} = 100 \, \text{volts}, \quad R = 800 \, \Omega\)
Substituting the values:
\(P = \frac{100^2}{800} = \frac{10000}{800} = 12.5 \, \text{watts}\)
Thus, the power consumed is:
\(P = 12.5 \, \text{watts}\)
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).