P(load) = I^2*Rload = 2.0 W
Rload = 2ohm
E = E/(Rload + Rinternal)
P = E^2*2/(2 + 0.1)= 0.9524 E^2 = 2
E = 1.449 V
a battery havibg an emf of E volt and internal resitance .1ohm is connected across the treminals A and B. Calculate the value of E inorder that the power dissipated in the 2 ohm resistoe is 2 watt
3 answers
P = V^2 / R = 2W.
V^2 / 2 = 2,
V^2 = 4,
V = 2 volts = voltage across 2 Ohms.
I = V / R = 2/2 = 1 Amp.
E = I * Rt = = 1*(0.1+2) = 2.1 Volts.
V^2 / 2 = 2,
V^2 = 4,
V = 2 volts = voltage across 2 Ohms.
I = V / R = 2/2 = 1 Amp.
E = I * Rt = = 1*(0.1+2) = 2.1 Volts.
Henry is correct. My answer was wrong
You need a current of I = 1 A to dissipate 2W in the 2 ohm resisitor.
I^2*R = 2
Total E of battery = I*(R + Ri)
= 1*(2+.1) = 2.1 V
You need a current of I = 1 A to dissipate 2W in the 2 ohm resisitor.
I^2*R = 2
Total E of battery = I*(R + Ri)
= 1*(2+.1) = 2.1 V
4 answers