Question

A 2.9 kΩ and a 3.8 kΩ resistor are connected in parallel; this combination is connected in series with a 1.2 kΩ resistor. If each resistor is rated at 0.7 W (maximum without overheating), what is the maximum voltage that can be applied across the whole network?

Answer 1

Rp = 2.9 x 3.8 / (2.9 + 3.8) = 1.64 k Ohm

Req = Rp + 1.2 = 2.84 k

I = V/Req = V/2.84 k Ohm

we know that the same current flows through the resistors connected in series.So the battery current will start and flow through 1.2 k Ohm first and then goes to parallel combination.

The drop across 1.2 is

V(1.2) = R(1.2) x I = 1.2 x V/2.84 = 0.42 V

we know that, Power dissipation is given by

P = I^R for R = 1.2 k Ohm

0.7 = (V/2.84)^2 x 1.2 = (V^2/8.07 x 10^6) x 1.2 x 10^3 = V^2 (0.149 x 10^-3)

V = 68.5 Volts