Question

Find the power delivered by the sources in the circuit shown in figure:

Answer 1

Hint: In question such as these we can easily see that this circuit form a closed loop so the best and the simplest way to solve his problem is by applying the Kirchhoff’s voltage law as a close loop is given, we can determine the unknown variables that helps in simplifying the problem. Kirchhoff’s voltage law can be applied anywhere if a closed loop is given.

Formula used:
$P = {i^2}R$
Where, $P$ is the power, $i$ is the current and $R$ is the resistance
Conventions:
-If moving in the direction of current voltage decreases if a resistor is encountered.
-Voltage decreases if the current moves against the direction of the voltage source(i.e., from – to +).

Complete step by step answer:
Let’s start by applying Kirchhoff’s voltage law in the loop ‘a b c d’, starting from the point a and moving in the direction of the current.

At first we encounter a voltage source of 20-volt but it is in the opposite direction to the flow of current so, as per the sign convention we get a decrease of 20-volt (-20 V). Then a resistor of 2 ohm, which decreases the potential by $2i$. A 50-volt voltage source increases the potential followed by the decrease of potential by $i$. Thus, we reach the point ‘a’ from where we started.

By Kirchhoff’s voltage law
$ - 20 - 2i + 50 - i = 0$
$\Rightarrow 30 - 3i = 0$
$\Rightarrow 30 = 3i$
$\Rightarrow i = 10$
Since we know the current, we can find the value of power dissipated by the formula $\left( {P = {i^2}R} \right)$
Total power:
$P = {10^2} \times 2 + {10^2} \times 1$
$\Rightarrow P = 200 + 100$
$\therefore P = 300\,W$

Thus, the total power dissipated is 300 watts.

Note: There is also a direct way to solve this problem for competitive exams.First, we find the total equivalent voltage of the circuit which is 30-volt as the two sources are align against each other, but the resistance gets added simply and these two are in series with each other, so total resistance becomes 3 ohms. Thus, we can find the current and solve this problem.