Question

What is the total power drawn by the circuit shown?

Answer 1

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:

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$

$\dfrac{1}{R_{eq}}=\dfrac{1}{20}+\dfrac{1}{8}+\dfrac{1}{500}$  

$R_{eq}=5.64\ \Omega$

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 :

$I=\dfrac{V}{R}$

$I=\dfrac{110\ V}{5.64\ \Omega}$

I = 19.5 A

Power drawn by the circuit is the product of the voltage and the current flowing i.e.

$P=V\times I$

$P=110\times 19.5$

P = 2145 Watts

or

P = 2.14 kW

Hence, this is the required solution.

Answer 2

The answer is 2.14 kW.

You get this by dividing the outside number (110) by all the resistors and then adding the values together.