A 20v battery with an internal resistance of 5ohm is connected to a resistance of x ohm if an additional 6ohm resistance is connected across the battery find the value of x so that the power supplied externally by the battery remains the same
To find the value of x which will keep the power supplied by the battery constant, we can use the formula for power:
P = (V^2) / R
Here:
V = 20V (battery voltage)
R = x (resistance connected across the battery)
We can first find the initial power supplied by the battery without the additional resistance:
P_initial = (V^2) / R_initial = (20^2) / x
Next, when an additional 6-ohm resistance is connected across the battery, the total resistance becomes:
R_total = R_initial + 6
To keep the power constant, we need to find the new value of x that satisfies:
P_initial = P_new
So,
(20^2) / x = (20^2) / (x + 6)
To solve this equation, we can cross-multiply:
(20^2) * (x + 6) = (20^2) * x
Simplifying:
(x + 6) = x
x + 6 - x = 0
6 = 0
Since the equation doesn't hold true, we can conclude that there is no value of x that will keep the power supplied by the battery constant when a 6-ohm resistance is added.
To find the value of x so that the power supplied externally by the battery remains the same, we need to calculate the power before and after connecting the additional 6-ohm resistance.
Let's start by calculating the initial power supplied by the battery before connecting the 6-ohm resistance.
Power (P) is given by the formula: P = (V^2) / R, where V is the voltage and R is the resistance.
Given:
Voltage (V) = 20V
Internal Resistance (r) = 5Ω
Total resistance before connecting the 6-ohm resistance:
R_total = x + internal resistance = x + 5Ω
Power before connecting the 6-ohm resistance:
P_initial = (V^2) / R_total
= (20^2) / (x + 5)
Now, let's calculate the power after connecting the 6-ohm resistance.
Total resistance after connecting the 6-ohm resistance:
R_total = x + 5Ω (internal resistance) + 6Ω (additional resistance)
= x + 11Ω
Power after connecting the 6-ohm resistance:
P_final = (V^2) / R_total
= (20^2) / (x + 11)
Since we want the power to remain the same: P_initial = P_final
Therefore, we can set up the equation:
(20^2) / (x + 11) = (20^2) / (x + 5)
Now, we can solve for x.
By cross-multiplication:
(20^2) * (x + 5) = (20^2) * (x + 11)
Simplifying:
(x + 5) = (x + 11)
Now, subtracting x from both sides:
5 = 11
Since this equation is not valid, it means there is no value of x that would make the power supplied externally by the battery remain the same.
Io = 20/(5+6) = 1.82A. = Output current.
Po = (1.82)^2 * 6 = 19.83 Watts.
Io^2 * Rt = 19.83.
(1.82)^2 * Rt = 19.83,
Rt = 2.06 Ohms.
6*X/(6+X) = 2.06.
6X = 12.36 + 2.06X,
3.94X = 12.36, X = 3.14 Ohms.