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
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
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.
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
To calculate the value of the electromotive force (EMF) required to have a power dissipation of 2 watts in a 2-ohm resistor connected to a battery with an internal resistance of 0.1 ohms, we can use the formula:
Power = (EMF^2 * R) / (R + r),
Where:
Power is the power dissipated in the resistor (2 watts),
EMF is the electromotive force of the battery (unknown),
R is the resistance of the resistor (2 ohms),
r is the internal resistance of the battery (0.1 ohms).
Rearranging the formula to solve for EMF, we get:
EMF = sqrt((Power * (R + r)) / R).
Substituting the given values into the equation:
EMF = sqrt((2 * (2 + 0.1)) / 2).
Simplifying further:
EMF = sqrt((2 * 2.1) / 2).
EMF = sqrt(4.2).
EMF ≈ 2.05 volts.
Therefore, the value of the electromotive force (E) required is approximately 2.05 volts.