Question

Draw a circuit diagram of an electric circuit containing a cell, a key, a resistor of 4$\Omega$ in series with a combination of two resistors (8$\Omega$ each) in parallel and key, an ammeter, a voltmeter across the parallel combination. Each of them dissipates maximum energy and can withstand a maximum power of 16W without melting. Find the maximum current that can flow through the three resistors.

Answer 1

We need to understand the relation between the type of network combination, the power dissipation in the circuit, and the current through the resistors involved in the circuit. We can solve them easily by using these parameters and relations.

Complete step-by-step solution
We can draw the diagram for the given set of resistors, key, ammeter, and voltmeter as shown below.

We can find the power dissipated in the circuit can be found by finding the equivalent resistance of the resistor combinations. The equivalent resistance of the circuit is given by finding the resistance offered by the parallel combination of the two resistors of 8 $\Omega$ each and the third resistor with the 4 $\Omega$ resistor.
Now, we are given the maximum power dissipation possible for each resistor to be 16W. The current that can pass through a resistor of 4 $\Omega$ resistance and power rating of 16W is –

& P={{I}^{2}}R \\\ & \Rightarrow I=\sqrt{\dfrac{P}{R}} \\\ & \Rightarrow I=\sqrt{\dfrac{16}{4}} \\\ & \therefore I=2A \\\ \end{aligned}$$ The current that can pass through a resistor of 8 $$\Omega $$ resistance and power rating of 16 W is – $$\begin{aligned} & P={{I}^{2}}R \\\ & \Rightarrow I=\sqrt{\dfrac{P}{R}} \\\ & \Rightarrow I=\sqrt{\dfrac{16}{8}} \\\ & \therefore I=\sqrt{2}A \\\ \end{aligned}$$ Now, we know that the maximum current through the circuit can be 2A. This current splits into half when it passes through the parallel combination of equal resistance. So, the current through the 8 $$\Omega $$ resistors will be 1 A each. So, the maximum current that can flow through the circuit is 2 A. The current through the 4 $$\Omega $$ resistor will be 2 A. The current through the 8 $$\Omega $$ resistor will be 1 A. This is the required solution. **Note:** In this case, the maximum current that can flow through the resistor of 4 $$\Omega $$ is 2 A, which can be easily divided among the two resistors of 8 $$\Omega $$ resistors. In other cases, we should take care to consider the current values in both combinations to avoid errors.