8. The current in a simple series circuit is 5 amp. When an additional resistance of 2 ohms is inserted in series, the current drops to 4 amp. What is the resistance of the original circuit?
V/r=5
V/(r+2)=3
V=5r
V=3r+6
5r=3r+6
r=3
To find the resistance of the original circuit, we can use Ohm's Law, which states that the current (I) flowing through a circuit is directly proportional to the voltage (V) and inversely proportional to the resistance (R).
We are given that the current in the original circuit is 5 amps, and when an additional resistance of 2 ohms is inserted, the current drops to 4 amps.
Let's denote the resistance of the original circuit as R1. The resistance we added is 2 ohms, so the total resistance of the circuit with the added resistance is R1 + 2.
According to Ohm's Law: I = V / R
Using this formula, we can set up two equations based on the given information:
Equation 1: 5 = V / R1
Equation 2: 4 = V / (R1 + 2)
Now, let's solve these equations simultaneously to find the value of R1.
From Equation 1:
5 x R1 = V
Substitute this value of V into Equation 2:
4 = (5 x R1) / (R1 + 2)
Cross-multiply to get rid of the fractions:
4(R1 + 2) = 5 x R1
Expand and simplify:
4R1 + 8 = 5R1
Rearrange the equation:
5R1 - 4R1 = 8
Combine like terms:
R1 = 8
Therefore, the resistance of the original circuit (R1) is 8 ohms.