The current a series circuit in 5A when an additional resistance of 2 ohm is inserted the current drops to 4A voltage remaining same the original resistance of the circuit is
original is r
v = 5 r
then later
same v = 4 (r+2)
so
5 r = 4 r + 8
r = 8 ohms
To find the original resistance of the circuit, we can use Ohm's Law, which states that the current passing through a conductor is directly proportional to the voltage and inversely proportional to the resistance.
In this case, we have a series circuit, where the total resistance is the sum of individual resistances. Let's denote the original resistance of the circuit as R.
According to the problem, when the circuit has only the original resistance, the current is 5A. And when an additional resistance of 2 ohms is added, the current drops to 4A. The voltage across the circuit remains the same.
Now, let's apply Ohm's Law to both scenarios:
1. Original circuit:
Current (I1) = 5A
Resistance (R) = ?
Voltage (V) = Constant
Using Ohm's Law: V = I1 * R
2. Circuit with an additional resistance:
Current (I2) = 4A
Total Resistance (R + 2Ω) = ?
Voltage (V) = Constant (same as in the original circuit)
Using Ohm's Law: V = I2 * (R + 2Ω)
Since V is constant in both scenarios, we can equate the two equations:
I1 * R = I2 * (R + 2)
Substituting the given values:
5A * R = 4A * (R + 2Ω)
Simplifying the equation:
5R = 4R + 8
Subtracting 4R from both sides:
R = 8
Therefore, the original resistance of the circuit is 8 ohms.