Current Electricity • NEET Revision

Example 3.49

Power Rating of Resistors

Question

In the following figure, each of the three resistances has a rating of 24 W and resistance of 6 Ω. Find the maximum power rating of the circuit. (Two 6 Ω resistors in parallel between A and B, in series with a 6 Ω resistor from B to C.)

Given

Each resistor: \(R = 6\ \Omega\), maximum safe power \(P_{max} = 24\ \text{W}\)
Circuit: \((6\ \Omega \parallel 6\ \Omega)\) from \(A\) to \(B\), in series with \(6\ \Omega\) from \(B\) to \(C\)

Asked

Maximum power the whole circuit can safely dissipate.

Formula

Safe current per resistor: \(I_{max} = \sqrt{\dfrac{P_{max}}{R}}\)

\(P_{circuit} = I^2 R_{total}\)

Solution

Safe current through any single resistor: \(I_{max} = \sqrt{\dfrac{24}{6}} = 2\ \text{A}\).
Total resistance: \(R_{AB} = \dfrac{6 \times 6}{6 + 6} = 3\ \Omega;\quad R_{total} = 3 + 6 = 9\ \Omega\).
The series resistor (B–C) carries the full circuit current \(I\), so it is the bottleneck: \(I \le 2\ \text{A}\).
At \(I = 2\ \text{A}\), each parallel resistor carries only \(1\ \text{A}\) \(\Rightarrow\) power \(= 1^2 \times 6 = 6\ \text{W} \le 24\ \text{W}\) — safe.
Maximum circuit power: \(P = I^2 R_{total} = 2^2 \times 9 = 36\ \text{W}\).
Check: series resistor \(2^2 \times 6 = 24\ \text{W}\) (at its limit) \(+\ 6 + 6\ \text{W}\) in the parallel pair \(= 36\ \text{W}\ \checkmark\)

\(P_{max} = 36\ \text{W}\)

Short Cut

Always hunt for the resistor carrying the largest current (or largest voltage) — it sets the limit. In a series–parallel mix, the lone series resistor is almost always the bottleneck. Here it hits 24 W first, and the rest of the circuit adds \(\tfrac{P}{2}\) more \(\Rightarrow 24 + 12 = 36\ \text{W}\).