To find the electric energy consumed by a 60 W bulb operating for 10 hours a day, we need to use the formula for electrical energy consumption:
\[ \text{Energy (kWh)} = \frac{\text{Power (W)} \times \text{Time (hours)}}{1000} \]
Given that the power of the bulb is 60 W and the time it operates per day is 10 hours, we substitute these values into the formula:
\[ \text{Energy} = \frac{60 \times 10}{1000} \]
\[ \text{Energy} = \frac{600}{1000} \]
\[ \text{Energy} = 0.6 \text{ kWh} \]
Thus, the electric energy consumed in operating the bulb for 10 hours a day is 0.6 kWh.
We are given:
Now, electric energy consumed is given by:
\[ \text{Energy (in kWh)} = \text{Power (in kW)} \times \text{Time (in hours)} \]
\[ = 0.06 \times 10 = 0.6 \, \text{kWh} \]
Correct Answer: 0.6
The maximum percentage error in the equivalent resistance of two parallel connected resistors of 100 \( \Omega \) and 900 \( \Omega \), with each having a maximum 5% error, is: \[ {(round off to nearest integer value).} \]
The induced emf in a 3.3 kV, 4-pole, 3-phase star-connected synchronous motor is considered to be equal and in phase with the terminal voltage under no-load condition. On application of a mechanical load, the induced emf phasor is deflected by an angle of \( 2^\circ \) mechanical with respect to the terminal voltage phasor. If the synchronous reactance is \( 2 \, \Omega \), and stator resistance is negligible, then the motor armature current magnitude, in amperes, during loaded condition is closest to: \[ {(round off to two decimal places).} \]
In the Wheatstone bridge shown below, the sensitivity of the bridge in terms of change in balancing voltage \( E \) for unit change in the resistance \( R \), in mV/\(\Omega\), is _____________ (round off to two decimal places).