Question

Three resistors 4Ω, 6Ω and 8Ω are connected in series to a battery terminal voltage of 12 volts. Calculate the current in the circuit, potential difference across each resistor and power dissipated in each resistor?​

Answer 1

Solution:-

Given:-

4Ω, 6Ω and 8Ω are connected in series.

∴Total resistance=(4+6+8)Ω=18Ω

I=V/R

I=12/18

I=2/3A

Vat 4Ω=I(R)

Vat 4 ohm=2/3(4)

V at 4 ohm=8/3V

Vat 6Ω=I(R)

Vat 6 ohm=2/3(6)

Vat 6 ohm=4V

Vat 8Ω=I(R)

V at 8 ohm=2/3(8)

V at 8 ohm=16/3V

P at 4Ω=V(I)

P at 4 ohm=8/3(2/3)

P at 4 ohm=16/9W

P at 6 ohm=V(I)

P at 6 ohm=4(2/3)

P at 6 ohm=8/3W

P at 8 ohm=V(I)

P at 8 ohm=16/3(2/3)

P at 8 ohm=32/9W