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

Question

Philoid · Accepted Answer

Here P =18 W

As resistor A is in series with the parallel combination of resistors B and C, so it carries maximum current.

We know P = I²R

I² = 9

I = 3 A

If I_B and I_C are current flowings through resistor B and C respectively and the potential difference will be same across both resistors B and C as they are connected in parallel.

So we know V = IR

I_BR_B = I_CR_C

So

I_B = I_C

But I = I_B + I_C

3 = 2I_B [ since I_B = I_C]

I_B = 3/2

I_B = 1.5 A

I_C = 1.5 A

So maximum current flows through resistor A is 3 A, B is 1.5 A and C is 1.5 A.