Five identical resistors, each of resistance R, are connected in series to a battery, and a total current of 9.74 A flows in the circuit.
i. Now, another resistor of resistance R is added to the circuit, in series with the rest of the resistors. Find the new total current in the circuit. I =
ii. In the original circuit: suppose another resistor of resistance R is added to the circuit, in parallel with the battery. Find the new total current in the circuit. I =
See previous post: Wed, 10-8-14, 7:55PM
To solve this problem, we'll first analyze the original circuit and then consider the modifications.
Original Circuit:
In series connection, the total resistance (R_total_s) is the sum of the resistances of individual resistors. Since all the resistors are identical, the total resistance is given by:
R_total_s = R + R + R + R + R = 5R
The total current (I_total_s) in a series circuit is the same at every point. The total current in the circuit is given as 9.74 A.
i. Circuit with an additional resistor in series:
When another resistor of resistance R is added in series, the total resistance becomes:
R_total_s_new = 5R + R = 6R
To find the new total current (I_total_s_new), we can use Ohm's Law, which states that the current is equal to the voltage divided by the resistance:
I = V / R
Since the battery voltage remains the same, we can write the equation for the original circuit and the new circuit as:
I_total_s = V / R_total_s (Equation 1)
I_total_s_new = V / R_total_s_new (Equation 2)
Dividing Equation 2 by Equation 1, we get:
I_total_s_new / I_total_s = (V / R_total_s_new) / (V / R_total_s)
= (R_total_s / R_total_s_new)
Substituting the values:
I_total_s_new / 9.74 A = (5R) / (6R)
Simplifying further:
I_total_s_new = (5/6) * 9.74 A
Therefore, the new total current in the circuit with the additional resistor in series is:
I_total_s_new = 8.12 A
ii. Circuit with an additional resistor in parallel:
When another resistor of resistance R is added in parallel with the rest of the circuit, the total resistance decreases.
In parallel connection, the total resistance (R_total_p) is given by:
1 / R_total_p = 1 / R + 1 / R + 1 / R + 1 / R + 1 / R
= 5 / R
Therefore, the new total resistance (R_total_p_new) after adding an additional resistor in parallel is:
1 / R_total_p_new = 1 / R_total_p + 1 / R
= (5 / R) + 1 / R
= 6 / R
Hence, the new total resistance becomes:
R_total_p_new = R / 6
To find the new total current (I_total_p_new), we can again use Ohm's Law. However, in a parallel circuit, the total current is the sum of the currents flowing through each branch.
Since the resistors are identical, the current through each resistor is the same. Let's call it I_individual.
I_total_p_new = I_individual + I_individual + I_individual + I_individual + I_individual + I / R
Since the battery voltage remains the same, we can write the equation for the original circuit and the new circuit as:
I_total_p = V / R_total_p (Equation 3)
I_total_p_new = V / R_total_p_new (Equation 4)
Dividing Equation 4 by Equation 3, we get:
(I_individual + I_individual + I_individual + I_individual + I_individual + I / R) / (V / R_total_p_new) = (I_individual / (V / R_total_p))
Simplifying further:
(I_individual / (V / (R / 6))) = ((5/6) * I) / (V / (5R))
Cancelling out common terms, we get:
I_individual * 6 = (5/6) * I
Simplifying further:
I_individual = (5/6) * I / 6
Therefore, the new total current in the circuit with the additional resistor in parallel is:
I_total_p_new = (5/6) * I / 6 + I / R
I_total_p_new = (5/36) * I + I / R
Substituting the given values:
I_total_p_new = (5/36) * 9.74 A + 9.74 A / R
Therefore, the new total current in the circuit with the additional resistor in parallel is:
I_total_p_new = (0.13528 + 9.74 / R) A