Question

Three resistors A, B, and C of 2$\Omega$ resistance is as shown in figure. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors.

Answer 1

We are given the power ratings of three resistors each with the same resistance. We have to find the maximum current that can be flow through each element using this circuit which is forming a network of series-parallel combinations.

Complete answer:
We have to find the maximum current that can flow through this circuit. It is already given that the maximum power is dissipated by the resistors. We can find the maximum current that can flow through the resistors without melting them using this power rating.
We know from the Ohm’s law that –

& V=IR \\\ & \Rightarrow \text{ }I=\dfrac{V}{R} \\\ \end{aligned}$$ We also know that the power dissipated can be expressed in terms of the voltage drop and the current flow. The power dissipation is given as – $$P=VI$$  We can substitute the voltage term in the above equation with that we know from Ohm's law. We will get a equation that relates power and the current as – $$\begin{aligned} & P=(IR)I \\\ & \Rightarrow P={{I}^{2}}R \\\ \end{aligned}$$ Now, we can find the current that a resistance can allow to pass without melting for a given wattage rating – It is already given – $$\begin{aligned} & R=2\Omega \\\ & \text{and,} \\\ & P=18W \\\ & P={{I}^{2}}R \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow \text{ }I=\sqrt{\dfrac{P}{R}} \\\ & \Rightarrow \text{ }I=\sqrt{\dfrac{18}{2}}=3A \\\ \end{aligned}$$ So, we understand that the maximum current that can pass through any of the resistors A, B and C is 3A. Now, we have to consider the given circuit, we know that the resistors B and C are in parallel connection. So, the current that will pass through them will be half the half current through A. The maximum current is 3A. Therefore, the current I that can pass through the circuit is 3A. **The current through the resistors B and C in this circuit will be 1.5A.** **Note:** In household connections, we use a parallel network to tackle this issue. The voltage from the source will remain a constant and the current can be adjusted according to the device for its own safety. The current is distributed in the circuit according to the load.