Answer:
Power, P = 2.14 kW
Explanation:
In the given figure, the resistors R₁ , R₂ and R₃ are in parallel. So, their equivalent resistance will be:
[tex]\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}[/tex]
[tex]\dfrac{1}{R_{eq}}=\dfrac{1}{20}+\dfrac{1}{8}+\dfrac{1}{500}[/tex]
[tex]R_{eq}=5.64\ \Omega[/tex]
So, the equivalent resistance of the circuit is 5.64 ohms.
It is also given that the potential difference is 110 V. The current flowing through the circuit can be calculated from the Ohm's law :
[tex]I=\dfrac{V}{R}[/tex]
[tex]I=\dfrac{110\ V}{5.64\ \Omega}[/tex]
I = 19.5 A
Power drawn by the circuit is the product of the voltage and the current flowing i.e.
[tex]P=V\times I[/tex]
[tex]P=110\times 19.5[/tex]
P = 2145 Watts
or
P = 2.14 kW
Hence, this is the required solution.
