A battery is connected to a resistance causing a current of 0.5A in the circuit. The current drops to 0.4A when additional resistance of 5 ohm is connected in series current will drop to 0.2A when the resistance is Ans:30ohms explain how?

Question

Answer

Solution:

Given data:
Initial current, I1 = 0.5 A
New current, I2 = 0.4 A
Additional resistance, R = 5 ohm
Current with additional resistance, I3 = 0.2 A

Using Ohm's law, we know that V = IR, where V is the potential difference, I is the current, and R is the resistance.

Calculating initial resistance:
V = I1 * R1
V = 0.5 A * R1
R1 = V / I1

Calculating new resistance:
V = I2 * (R1 + R)
V = 0.4 A * (R1 + 5 ohm)
R1 + 5 ohm = V / 0.4 A
R1 + 5 ohm = 2.5 V
R1 = 2.5 V - 5 ohm = 0.5 V / 0.5 A = 1 ohm

Calculating resistance with current of 0.2 A:
V = I3 * (R1 + R2)
V = 0.2 A * (1 ohm + R2)
1 ohm + R2 = V / 0.2 A
1 ohm + R2 = 5 V
R2 = 5 V - 1 ohm = 4 V / 0.2 A = 20 ohm

Therefore, the resistance needed to decrease the current from 0.5 A to 0.2 A is 20 ohm.

Explanation:

The initial current in the circuit is 0.5 A. When additional resistance of 5 ohm is connected in series, the current drops to 0.4 A. This means that the total resistance of the circuit has increased.

Using Ohm's law, we can calculate the initial resistance of the circuit. Then, using the fact that the potential difference across the battery remains constant, we can calculate the new resistance of the circuit when the additional resistance is added. We can then use Ohm's law again to calculate the resistance needed to reduce the current to 0.2 A.

Overall, the calculations show that as the resistance of the circuit increases, the current decreases, and vice versa.

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