In the circuit given below, each of three resistors of \(4\Omega\) can have a maximum power of \(20\,\text{W}\) otherwise, it will melt. The maximum power which the whole circuit can take is:
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The resistor carrying highest current usually reaches limiting power first. Check that resistor before calculating total allowable power.
Step 1: Understand the safety limit. Each $4\,\Omega$ resistor can take at most $20$ W before it melts. The resistor that carries the most current is the one that hits this limit first, so it sets the maximum safe power for the whole circuit.
Step 2: Reduce the network. Two of the $4\,\Omega$ resistors are in parallel. \[ R_p = \frac{4 \times 4}{4 + 4} = 2\,\Omega \] This is in series with the third $4\,\Omega$. \[ R_{eq} = 2 + 4 = 6\,\Omega \] Step 3: Find which resistor is most loaded. The full current passes through the series $4\,\Omega$ resistor, so it reaches its limit first. Step 4: Use the limit on that resistor. \[ P = I^2 R \quad\Rightarrow\quad 20 = I^2 (4) \] \[ I^2 = 5 \] Step 5: Find the total circuit power. The total power is the same current squared times the whole resistance. \[ P_{total} = I^2 R_{eq} = 5 \times 6 \] Step 6: State the maximum power. \[ P_{total} = 30\,\text{W} \] \[ \boxed{30\,\text{W}} \]
