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An inductor coil with a resistance of 10Ω and an inductance of 120 mH is connected across a battery with an emf of 6 V and an internal resistance of 2Ω. Find the charge that flows through the inductor in 10 ms.

Answer :

Final answer:

To calculate the charge flowing through an inductor in 10 ms, the time constant of the RL circuit is first determined. Applying the formula for current growth in an RL circuit, the current at 10 ms is found. Multiplying by time gives the charge, which is approximately 3.16 mC.

Explanation:

To calculate the charge that flows through the inductor in 10 ms, we first need to determine the time constant of the RL circuit. The time constant (τ) is given by the ratio of inductance (L) to resistance (R), which includes both the inductor's resistance and the internal resistance of the battery:

τ = L / (R + r)

Where:

  • L = 120 mH (0.120 H)
  • R = 10Ω (inductor's resistance)
  • r = 2Ω (battery's internal resistance)

This gives us:

τ = 0.120 H / (10Ω + 2Ω) = 0.120 H / 12Ω = 0.01 s

The current through the inductor in an RL circuit increases according to the formula:

I(t) = V/R * (1 - e^(-t/τ))

Where:

  • V is the emf of the battery (6V)
  • I(t) is the current at time t
  • t is the time (10 ms or 0.01 s)

Substituting our values, we get:

I(10 ms) = 6V / 12Ω * (1 - e^(-0.01s/0.01s))

I(10 ms) = 0.5 A * (1 - e^(-1))

Now, to find the charge (Q) that flows through, we use the formula:

Q = I(t) * t

Substituting the current at 10ms and the time, we have:

Q = 0.5 A * (1 - e^(-1)) * 0.01s

Q ≈ 0.5 A * 0.632 * 0.01s = 3.16 * 10^-3 C

Therefore, the charge which flows through the inductor in 10 ms is approximately 3.16 mC.

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