The correct answer is \(14W\)
\(R_1=\frac{(220)^2}{100}=484 Ω\), \(R_2=\frac{(220)^2}{60}=\frac{48400}{60} Ω\)
Current (i) 1=\(\frac{220}{484\frac{4840}{6}}=\frac{15}{88}amp\)
\(P_1=\left (\frac{15}{88}\right)^2×484=14.06W=14W\)
The correct answer is \(14W\)
\(R_1=\frac{V^2}{P}=\frac{220^2}{100}=484\)
\(R_2=\frac{V^2}{P}=\frac{220^2}{60}=484(\frac{10}{6})\)
\(I=\frac{220}{484+484\times\frac{10}{6}}\)
\(P_1=I^2R_1=14.06W=14W\)
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to:
Current electricity is defined as the flow of electrons from one section of the circuit to another.
There are two types of current electricity as follows:
The current electricity whose direction remains the same is known as direct current. Direct current is defined by the constant flow of electrons from a region of high electron density to a region of low electron density. DC is used in many household appliances and applications that involve a battery.
The current electricity that is bidirectional and keeps changing the direction of the charge flow is known as alternating current. The bi-directionality is caused by a sinusoidally varying current and voltage that reverses directions, creating a periodic back-and-forth motion for the current. The electrical outlets at our homes and industries are supplied with alternating current.